Network analysis by van valkenburg solutions CH#6 part 1

Muhammad Irfan Yousuf {Peon of Holy Prophet (P.B.U.H)} 1

CH#6


PROBLEMSQ#6.1: Show that i = ke-2t and i = ke-t are solutions of the differential equationd2i di + 3 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 3si + 2i = 0s2 + 3s + 2 = 0

a 1b 3c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -3 32 – 4(1)(2)s1, s2 = 2(1) -3 9 – 8s1, s2 = 2(1) -3 1s1, s2 = 2(1) -3 1s1, s2 =


2s1, s2 = (-3 + 1)/2, (-3 - 1)/2s1, s2 = (-2)/2, (-4)/2s1, s2 = -1, -2General solution:i(t) = K1es1t + K2es2t

Putting corresponding values we get

i(t) = K1e-1t + K2e-2t

Q#6.2: Show that i = Ke-t and i = Kte-t are solutions of the differential equationd2i di + 2 + i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Here1s2i + 2si + 1i = 0s2 + 2s + 1 = 0

a 1b 2c 1

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get

-2 22 – 4(1)(1)s1, s2 = 2(1) -2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 = 2s1, s2 = (-2 + 0)/2, (-2 - 0)/2s1, s2 = (-1), (-1)General solution:i(t) = K1es1t + K2tes2t



i(t) = K1e-1t + K2te-1t

Q#6.3: Find the general solution of each of the following equations:(a) d2i di + 3 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 3si + 2i = 0s2 + 3s + 2 = 0

a 1b 3c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -3 32 – 4(1)(2)s1, s2 = 2(1) -3 9 – 8s1, s2 = 2(1) -3 1s1, s2 = 2(1) -3 1s1, s2 = 2s1, s2 = (-3 + 1)/2, (-3 - 1)/2s1, s2 = (-2)/2, (-4)/2s1, s2 = -1, -2General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-1t + K2e-2t

(b)d2i di + 5 + 6i = 0dt2 dtCharacteristic equation:


as2 + bs + c = 0Heres2i + 5si + 6i = 0s2 + 5s + 6 = 0

a 1b 5c 6


-5 52 – 4(1)(6)s1, s2 = 2(1) -5 25 – 24s1, s2 = 2(1) -5 1s1, s2 = 2(1) -5 1s1, s2 = 2s1, s2 = (-5 + 1)/2, (-5 - 1)/2s1, s2 = (-4)/2, (-6)/2s1, s2 = -2, -3General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-2t + K2e-3t

(c)d2i di + 7 + 12i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 7si + 12i = 0s2 + 7s + 12 = 0

a 1


b 7c 12

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -7 72 – 4(1)(12)s1, s2 = 2(1) -7 49 – 48s1, s2 = 2(1)

-7 1s1, s2 = 2(1) -7 1s1, s2 = 2s1, s2 = (-7 + 1)/2, (-7 - 1)/2s1, s2 = (-6)/2, (-8)/2s1, s2 = -3, -4General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-3t + K2e-4t

(d)d2i di + 5 + 4i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 5si + 4i = 0s2 + 5s + 4 = 0

a 1b 5c 4

-b b2 – 4acs1, s2 = 2a





i(t) = K1e-1t + K2e-4t

(e)d2i di + 1 + 6i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 1si + 6i = 0s2 + 1s + 6 = 0

a 1b 1c 6



-1 -23s1, s2 = 2(1) -1 i4.796s1, s2 = 2s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t

(f)d2i di + 1 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 1si + 2i = 0s2 + 1s + 2 = 0

A 1B 1C 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -1 12 – 4(1)(2)s1, s2 = 2(1) -1 1 – 8s1, s2 = 2(1) -1 -7s1, s2 = 2(1) -1 i2.646s1, s2 = 2s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323)General solution:i(t) = K1es1t + K2es2t



i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t

(g)d2i di + 2 + i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 2si + i = 0s2 + 2s + 1 = 0

a 1b 2c 1

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2 22 – 4(1)(1)s1, s2 = 2(1) -2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 = 2s1, s2 = (-2 + 0)/2, (-2 - 0)/2s1, s2 = (-1), (-1)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-1)t + K2te(-1)t

(h)

d2i di + 4 + 4i = 0dt2 dtCharacteristic equation:


as2 + bs + c = 0Heres2i + 4si + 4i = 0s2 + 4s + 4 = 0

a 1b 4c 4



i(t) = K1e(-2)t + K2te(-2)t

Q#6.4: Find the general solution of each of the following homogeneous differential equations:(a)d2v dv + 2 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 2v = 0s2 + 2s + 2 = 0

a 1b 2


c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2 22 – 4(1)(2)s1, s2 = 2(1)

-2 4 – 8s1, s2 = 2(1) -2 -4s1, s2 = 2(1) -2 i2s1, s2 = 2s1, s2 = (-2 + i2)/2, (-2 – i2)/2s1, s2 = (-1 + i1), (-1 – i1)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t

(b)d2v dv + 2 + 4v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 4v = 0s2 + 2s + 4 = 0

a 1b 2c 4

-b b2 – 4acs1, s2 = 2a



-2 22 – 4(1)(4)s1, s2 = 2(1) -2 4 – 16s1, s2 = 2(1)

-2 -12s1, s2 = 2(1) -2 i3.464s1, s2 = 2s1, s2 = (-2 + i3.464)/2, (-2 – i3.464)/2s1, s2 = (-1 + i1.732), (-1 – i1.732)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t

(c)d2v dv + 4 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 4sv + 2v = 0s2 + 4s + 2 = 0

a 1b 4c 2


-4 42 – 4(1)(2)s1, s2 =


2(1) -4 16 – 8s1, s2 = 2(1) -4 8s1, s2 = 2(1) -4 2.828s1, s2 = 2s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2s1, s2 = (-0.586), (-3.414)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-0.586)t + K2e(-3.414)t

(d) d2v dv2 + 8 + 16v = 0 dt2 dtAfter simplificationd2v dv + 4 + 8v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 4sv + 8v = 0s2 + 4s + 8 = 0

a 1b 4c 8


-4 42 – 4(1)(8)s1, s2 = 2(1) -4 16 – 32s1, s2 = 2(1)


-4 -16s1, s2 = 2(1) -4 i4s1, s2 = 2s1, s2 = (-4 + i4)/2, (-4 – i4)/2s1, s2 = (-2 + i2), (-2 – i2)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t

(e)d2v dv + 2 + 3v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 3v = 0s2 + 2s + 3 = 0

a 1b 2c 3


-2 22 – 4(1)(3)s1, s2 = 2(1) -2 4 – 12s1, s2 = 2(1) -2 -8s1, s2 = 2(1) -2 i2.828s1, s2 = 2s1, s2 = (-2 + i2.828)/2, (-2 – i2.828)/2s1, s2 = (-1 + i1.414), (-1 – i1.414)


General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t

(f)d2v dv + 3 + 5v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 3sv + 5v = 0s2 + 3s + 5 = 0

a 1b 3c 5


-3 32 – 4(1)(5)s1, s2 = 2(1) -3 9 – 20s1, s2 = 2(1) -3 -11s1, s2 = 2(1) -3 i3.317s1, s2 = 2s1, s2 = (-3 + i3.317)/2, (-3 – i3.317)/2s1, s2 = (-1.5 + i1.659), (-1.5 – i1.659)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t


Q#6.5: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: dii(0+) = 1, (0+) = 0 dt(a)d2i di + 3 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 3si + 2i = 0s2 + 3s + 2 = 0

a 1b 3c 2



i(t) = K1e-1t + K2e-2t … (i)

At t = 0+i(0+) = K1e-1(0+) + K2e-2(0+)


1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’

di(t) = -K1e-1t - 2K2e-2t

dtdi(0+) = -K1e-1(0+) - 2K2e-2(0+)

dtdi(0+) = -K1e0 - 2K2e0

dt0 = -K1(1) - 2K2(1)0 = -K1 - 2K2

K1 = - 2K2

Putting the value of K1 in (ii)1 = -2K2 + K2

K2 = -1Putting the value of K2 in (ii)1 = K1 + (-1)K1 = 2 Putting corresponding values we geti(t) = 2e-t + (-1)e-2t

i(t) = 2e-t – e-2t

(b)d2i di + 5 + 6i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 5si + 6i = 0s2 + 5s + 6 = 0

A 1B 5C 6

-b b2 – 4ac


s1, s2 = 2aPutting corresponding values we get



i(t) = K1e-2t + K2e-3t … (i)

At t = 0+i(0+) = K1e-2(0+) + K2e-3(0+)

i(0+) = K1e0 + K2e0

i(0+) = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’di(t) = -2K1e-2t - 3K2e-3t

dtdi(0+) = -2K1e-2(0+) - 3K2e-3(0+)

dtdi(0+) = -2K1e0 - 3K2e0

dt

di(0+) = -2K1(1) - 3K2(1)


dt0 = -2K1 - 3K2

2K1 = -3K2 K1 = -1.5K2

1 = K1 + K2 … (ii)Putting the value of K1 in (ii)1 = -1.5K2 + K2

1 = -0.5K2 K2 = -2K1 = -1.5(-2)K1 = 3.0Putting corresponding values we geti(t) = 3e-2t + (-2)e-3t

i(t) = 3e-2t - 2e-3t


a 1b 7c 12


-7 1s1, s2 = 2(1) -7 1s1, s2 = 2s1, s2 = (-7 + 1)/2, (-7 - 1)/2


s1, s2 = (-6)/2, (-8)/2s1, s2 = -3, -4General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-3t + K2e-4t … (i)

At t = 0+i(0+) = K1e-3(0+) + K2e-4(0+)

i(0+) = K1e0 + K2e0


dtdi(0+) = -3K1e-3(0+) - 4K2e-4(0+)

dtdi(0+) = -3K1e0 - 4K2e0

dtdi(0+) = -3K1(1) - 4K2(1) dtdi(0+) = -3K1 - 4K2

dt0 = -3K1 - 4K2

3K1 = - 4K2

K1 = -1.334K2

Putting the value of K1 in (ii)1 = -1.334K2 + K2

1 = -0.334K2 K2 = -2.994 K1 = -1.334(-2.994) = 3.994Putting corresponding values we geti(t) = (3.994)e-3t + (-2.994)e-4t

i(t) = 3.994e-3t - 2.994e-4t

(d)d2i di + 5 + 4i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0


Heres2i + 5si + 4i = 0s2 + 5s + 4 = 0

a 1b 5c 4




i(t) = K1e-1t + K2e-4t … (i)At t = 0+i(0+) = K1e-1(0+) + K2e-4(0+)

i(0+) = K1e0 + K2e0


dtdi(0+) = -1K1e-1(0+) - 4K2e-4(0+)

dt


di(0+) = -1K1e0 - 4K2e0

dt0 = -1K1(1) - 4K2(1)0 = -K1 - 4K2

K1 = -4K2


1 = -3K2

K2 = -0.334K1 = -4(-0.334)K1 = 1.336Putting corresponding values we get

i(t) = 1.336e-1t – 0.334e-4t

(e)d2i di + 1 + 6i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 1si + 6i = 0s2 + 1s + 6 = 0

A 1B 1C 6


-1 12 – 4(1)(6)s1, s2 = 2(1) -1 1 – 24s1, s2 = 2(1) -1 -23s1, s2 = 2(1)

-1 i4.796


s1, s2 = 2s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t … (i)At t = 0+i(0+) = K1e(-0.5 + i2.398)0+ + K2e(-0.5 - i2.398)0+

i(0+) = K1e0 + K2e0

i(0+) = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’di(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t

dtdi(0+) = (-0.5 + i2.398)K1e(-0.5 + i2.398)0+ + (-0.5 - i2.398)K2e(-0.5 - i2.398)0+

dtdi(0+) = (-0.5 + i2.398)K1e0 + (-0.5 - i2.398)K2e0

dt0 = (-0.5 + i2.398)K1(1) + (-0.5 - i2.398)K2(1)0 = (-0.5 + i2.398)K1 + (-0.5 - i2.398)K2

0.5 + i2.398K1 = K2

-0.5 + i2.398K1 = [0.917 – i0.400]K2

Putting the value of K1 in (ii)1 = [0.917 – i0.400]K2 + K2 1 = [[0.917 – i0.400] + 1]K2

1 = [0.917 – i0.400 + 1]K2

1 = [1.917 – i0.400]K2

K2 = 0.5 + i0.1041 = K1 + K2

K1 = 1 – K2

K1 = 1 – [0.5 + i0.104]K1 = 1 – 0.5 - i0.104K1 = 0.5 - i0.104Putting corresponding values we get

i(t) = [0.5 - i0.104]e(-0.5 + i2.398)t + [0.5 + i0.104]e(-0.5 - i2.398)t

(f)d2i di


+ 1 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 1si + 2i = 0s2 + 1s + 2 = 0

a 1b 1c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -1 12 – 4(1)(2)s1, s2 = 2(1) -1 1 – 8s1, s2 = 2(1) -1 -7s1, s2 = 2(1) -1 i2.646s1, s2 = 2s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t … (i)

At t = 0+i(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 - i1.323)0+

i(0+) = K1e0 + K2e0

i(0+) = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’

di(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t


dt

di(0+) = (-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 - i1.323)K2e(-0.5 - i1.323)0+

dtdi(0+) = (-0.5 + i1.323)K1e0 + (-0.5 - i1.323)K2e0

dtdi(0+) = (-0.5 + i1.323)K1(1) + (-0.5 - i1.323)K2(1) dt0 = (-0.5 + i1.323)K1 + (-0.5 - i1.323)K2

(-0.5 + i1.323)K1 = (0.5 + i1.323)K2 (0.5 + i1.323)K1 = K2

(-0.5 + i1.323)K1 = (0.75 – 0.662i)K2

1 = (0.75 – 0.662i)K2 + K2

1 = [(0.75 – 0.662i) + 1]K2

1 = [1.75 – 0.662i]K2

K2 = 0.5 + 0.189iPutting corresponding values we get1 = K1 + 0.5 + 0.189i

K1 = 1 - 0.5 - 0.189iK1 = 0.5 - 0.189iPutting corresponding values we get

i(t) = [0.5 - 0.189i]e(-0.5 + i1.323)t + [0.5 + 0.189i]e(-0.5 - i1.323)t


a 1b 2c 1

-b b2 – 4acs1, s2 = 2a


Putting corresponding values we get -2 22 – 4(1)(1)s1, s2 = 2(1)

-2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 = 2s1, s2 = (-2 + 0)/2, (-2 - 0)/2s1, s2 = (-1), (-1)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-1)t + K2te(-1)t … (i)

At t = 0+i(t) = K1e(-1)t + K2te(-1)t

i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+

i(0+) = K1e0

i(0+) = K1(1)1 = K1

Differentiating (i) with respect to ‘t’di(t) = -K1e(-1)t + K2[-te(-1)t – e-t] dtdi(0+) = (-1)K1e(-1)0+ + K2[-(0+)e-(0+) – e-(0+)] dt

di(0+) = (-1)K1e0 + K2[–e0] dtdi(0+) = (-1)K1(1) + K2[–(1)] dt 0 = -K1 – K2

0 = -(1) – K2

K2 = -1


Putting corresponding values i(t) = (1)e(-1)t + (-1)te(-1)t

i(t) = e(-1)t - te(-1)t

(h)

d2i di + 4 + 4i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 4si + 4i = 0s2 + 4s + 4 = 0

A 1B 4C 4



i(t) = K1e(-2)t + K2te(-2)t

At t = 0+i(0+) = K1e(-2)0+ + K2(0+)e(-2)0+

1 = K1e0


1 = K1(1)1 = K1

Differentiating (i) with respect to ‘t’di(t) = -2K1e(-2)t + K2[-0.5te-2t – 0.25e-2t] dtdi(0+) = -2K1e(-2)0+ + K2[-0.5(0+)e-2(0+) – 0.25e-2(0+)] dtdi(0+) = -2K1e0 + K2[–0.25e0] dtdi(0+) = -2K1(1) + K2[–0.25(1)] dt0 = -2K1

– 0.25K2

2K1 = -0.25K2

K1 = -0.125K2

K2 = -8Putting corresponding valuesi(t) = (1)e(-2)t + (-8)te(-2)t

i(t) = e(-2)t - 8te(-2)t

Q#6.6: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: dii(0+) = 2, (0+) = +1 dt(a)d2i di + 3 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 3si + 2i = 0s2 + 3s + 2 = 0

a 1b 3c 2

-b b2 – 4acs1, s2 = 2a


Putting corresponding values we get -3 32 – 4(1)(2)s1, s2 = 2(1) -3 9 – 8s1, s2 = 2(1) -3 1s1, s2 = 2(1) -3 1s1, s2 = 2s1, s2 = (-3 + 1)/2, (-3 - 1)/2s1, s2 = (-2)/2, (-4)/2s1, s2 = -1, -2General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-1t + K2e-2t … (i)

At t = 0+i(0+) = K1e-1(0+) + K2e-2(0+)

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’di(t) = -K1e-1t - 2K2e-2t

dtdi(0+) = -K1e-1(0+) - 2K2e-2(0+)

dtdi(0+) = -K1e0 - 2K2e0

dt1 = -K1(1) - 2K2(1)1 = -K1 - 2K2

K1 = -1 - 2K2

Putting the value of K1 in (ii)2 = [-1 - 2K2] + K2

2 = -1 - 2K2 + K2

3 = -K2

K2 = -3Putting the value of K2 in (ii)


2 = K1 + (-3)K1 = 5 Putting corresponding values we geti(t) = 5e-t + (-3)e-2t

i(t) = 5e-t – 3e-2t

(b)d2i di + 5 + 6i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 5si + 6i = 0s2 + 5s + 6 = 0

A 1B 5C 6



i(t) = K1e-2t + K2e-3t … (i)


At t = 0+i(0+) = K1e-2(0+) + K2e-3(0+)

i(0+) = K1e0 + K2e0


dtdi(0+) = -2K1e-2(0+) - 3K2e-3(0+)

dtdi(0+) = -2K1e0 - 3K2e0

dtdi(0+) = -2K1(1) - 3K2(1) dt1 = -2K1 - 3K2

2K1 = -3K2 - 1 K1 = -1.5K2 – 0.52 = K1 + K2 … (ii)Putting the value of K1 in (ii)2 = [-1.5K2 – 0.5] + K2

2 = -1.5K2 – 0.5 + K2

2 = -0.5K2 – 0.5 K2 = -52 = K1 + (-5)K1 = 7Putting corresponding values we geti(t) = 7e-2t + (-5)e-3t

i(t) = 7e-2t - 5e-3t


a 1b 7c 12

-b b2 – 4ac


s1, s2 = 2aPutting corresponding values we get -7 72 – 4(1)(12)s1, s2 = 2(1) -7 49 – 48s1, s2 = 2(1)

-7 1s1, s2 = 2(1) -7 1s1, s2 = 2s1, s2 = (-7 + 1)/2, (-7 - 1)/2s1, s2 = (-6)/2, (-8)/2s1, s2 = -3, -4General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-3t + K2e-4t … (i)

At t = 0+i(0+) = K1e-3(0+) + K2e-4(0+)

i(0+) = K1e0 + K2e0


dtdi(0+) = -3K1e-3(0+) - 4K2e-4(0+)

dt

di(0+) = -3K1e0 - 4K2e0

dtdi(0+) = -3K1(1) - 4K2(1) dt


di(0+) = -3K1 - 4K2

dt1 = -3K1 - 4K2

3K1 = -4K2 - 1K1 = -1.334K2 – 0.334Putting the value of K1 in (ii)2 = [-1.334K2 – 0.334] + K2

2 = -1.334K2 – 0.334 + K2

2.334 = -0.334K2

K2 = -6.9882 = K1 + (-6.988)8.988 = K1 Putting corresponding values we geti(t) = (8.988)e-3t + (-6.988)e-4t

i(t) = 8.988e-3t – 6.988e-4t

(d)d2i di + 5 + 4i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 5si + 4i = 0s2 + 5s + 4 = 0

a 1b 5c 4


-5 52 – 4(1)(4)s1, s2 = 2(1) -5 25 – 16s1, s2 = 2(1)

-5 9


s1, s2 = 2(1) -5 3s1, s2 = 2s1, s2 = (-5 + 3)/2, (-5 - 3)/2s1, s2 = (-2)/2, (-8)/2s1, s2 = -1, -4General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-1t + K2e-4t … (i)At t = 0+i(0+) = K1e-1(0+) + K2e-4(0+)

i(0+) = K1e0 + K2e0


dtdi(0+) = -1K1e-1(0+) - 4K2e-4(0+)

dtdi(0+) = -1K1e0 - 4K2e0

dt0 = -1K1(1) - 4K2(1)0 = -K1 - 4K2

K1 = -4K2


1 = -3K2

K2 = -0.334K1 = -4(-0.334)K1 = 1.336Putting corresponding values we get

i(t) = 1.336e-1t – 0.334e-4t

(e)d2i di + 1 + 6i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0


Heres2i + 1si + 6i = 0s2 + 1s + 6 = 0

a 1b 1c 6


-1 12 – 4(1)(6)s1, s2 = 2(1) -1 1 – 24s1, s2 = 2(1) -1 -23s1, s2 = 2(1) -1 i4.796s1, s2 = 2s1, s2 = (-1 + i4.796)/2, (-1 - i4.796)/2s1, s2 = (-0.5 + i2.398), (-0.5 - i2.398)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-0.5 + i2.398)t + K2e(-0.5 - i2.398)t … (i)At t = 0+i(0+) = K1e(-0.5 + i2.398)0+ + K2e(-0.5 - i2.398)0+

i(0+) = K1e0 + K2e0


dtdi(0+) = (-0.5 + i2.398)K1e(-0.5 + i2.398)0+ + (-0.5 - i2.398)K2e(-0.5 - i2.398)0+

dtdi(0+) = (-0.5 + i2.398)K1e0 + (-0.5 - i2.398)K2e0

dt


1 = (-0.5 + i2.398)K1(1) + (-0.5 - i2.398)K2(1)1 = (-0.5 + i2.398)K1 + (-0.5 - i2.398)K2

After simplification K1 = -0.084 – i0.400 + [0.917 – i0.4]K2

2 = [-0.084 – i0.400 + [0.917 – i0.4]K2] + K2

2 + 0.084 + i0.400 = [1.917 – i0.4]K2

2.084 + i0.400 = [1.917 – i0.4]K2

K2 = 1 + i0.4172 = K1 + 1 + i0.4172 - 1 - i0.417 = K1 1 - i0.417 = K1


i(t) = [1 - i0.417]e(-0.5 + i2.398)t + [1 + i0.417]e(-0.5 - i2.398)t

(f)d2i di + 1 + 2i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 1si + 2i = 0s2 + 1s + 2 = 0

a 1b 1c 2


-1 12 – 4(1)(2)s1, s2 = 2(1) -1 1 – 8s1, s2 = 2(1) -1 -7s1, s2 = 2(1) -1 i2.646s1, s2 = 2


s1, s2 = (-1 + i2.646)/2, (-1 - i2.646)/2s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323)General solution:i(t) = K1es1t + K2es2t


i(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 - i1.323)t … (i)

At t = 0+i(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 - i1.323)0+

i(0+) = K1e0 + K2e0


dtdi(0+) = (-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 - i1.323)K2e(-0.5 - i1.323)0+

dtdi(0+) = (-0.5 + i1.323)K1e0 + (-0.5 - i1.323)K2e0

dtdi(0+) = (-0.5 + i1.323)K1(1) + (-0.5 - i1.323)K2(1) dt1 = (-0.5 + i1.323)K1 + (-0.5 - i1.323)K2

K1 = -0.25 – i0.662 + [0.75 – i0.662]K2

2 = -0.25 – i0.662 + [0.75 – i0.662]K2 + K2

2 + 0.25 + i0.662 = [0.75 – i0.662]K2 + K2

2.25 + i0.662 = [1.75 – i0.662]K2

K2 = 1 + i0.7572 = K1 + 1 + i0.757K1 = 2 – 1 - i0.757K1 = 1 - i0.757

i(t) = [1 - i0.757]e(-0.5 + i1.323)t + [1 + i0.757]e(-0.5 - i1.323)t



A 1B 2C 1




i(t) = K1e(-1)t + K2te(-1)t … (i)

At t = 0+i(t) = K1e(-1)t + K2te(-1)t

i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+

i(0+) = K1e0

i(0+) = K1(1)2 = K1

Differentiating (i) with respect to ‘t’di(t) = -K1e(-1)t + K2[-te(-1)t – e-t] dtdi(0+) = (-1)K1e(-1)0+ + K2[-(0+)e-(0+) – e-(0+)] dtdi(0+) = (-1)K1e0 + K2[–e0] dtdi(0+) = (-1)K1(1) + K2[–(1)]


dt 1 = -K1 – K2

1 = -(2) – K2

K2 = -3Putting corresponding values i(t) = (2)e(-1)t + (-3)te(-1)t

i(t) = 2e(-1)t - 3te(-1)t

(h)d2i di + 4 + 4i = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2i + 4si + 4i = 0s2 + 4s + 4 = 0

A 1B 4C 4





i(t) = K1e(-2)t + K2te(-2)t

At t = 0+i(0+) = K1e(-2)0+ + K2(0+)e(-2)0+

2 = K1e0

2 = K1(1)2 = K1

Differentiating (i) with respect to ‘t’di(t) = -2K1e(-2)t + K2[-0.5te-2t – 0.25e-2t] dtdi(0+) = -2K1e(-2)0+ + K2[-0.5(0+)e-2(0+) – 0.25e-2(0+)] dtdi(0+) = -2K1e0 + K2[–0.25e0] dtdi(0+) = -2K1(1) + K2[–0.25(1)] dt1 = -2K1

– 0.25K2

1 = -2(2) – 0.25K2

-5 = 0.25K2

K2 = -20Putting corresponding valuesi(t) = (2)e(-2)t + (-20)te(-2)t

i(t) = 2e(-2)t - 20te(-2)t

Q#6.7: Find particular solutions for the differential equations of Prob. 6.3 subject to the initial conditions: dvv(0+) = 1, (0+) = -1 dt(a)d2v dv + 2 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 2v = 0s2 + 2s + 2 = 0

a 1b 2


c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2 22 – 4(1)(2)s1, s2 = 2(1) -2 4 – 8s1, s2 = 2(1) -2 -4s1, s2 = 2(1) -2 i2s1, s2 = 2s1, s2 = (-2 + i2)/2, (-2 – i2)/2s1, s2 = (-1 + i1), (-1 – i1)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t … (i)

At t = 0+v(0+) = K1e(-1 + i1)0+ + K2e(-1 – i1)0+

v(0+) = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’dv(t) = (-1 + i1)K1e(-1 + i1)t + (-1 – i1)K2e(-1 – i1)t

dt

dv(0+) = (-1 + i1)K1e(-1 + i1)0+ + (-1 – i1)K2e(-1 – i1)0+

dtdv(0+) = (-1 + i1)K1e0 + (-1 – i1)K2e0

dtdv(0+) = (-1 + i1)K1(1) + (-1 – i1)K2(1) dt-1 = (-1 + i1)K1 + (-1 – i1)K2


K1 = -iK2 + 0.5 + 0.5i1 = [-iK2 + 0.5 + 0.5i] + K2

1 = -iK2 + 0.5 + 0.5i + K2

K2 = 0.51 = K1 + 0.5K1 = 0.5

v(t) = 0.5e(-1 + i1)t + 0.5e(-1 – i1)t

(b)d2v dv + 2 + 4v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 4v = 0s2 + 2s + 4 = 0

a 1b 2c 4




v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t … (i)


At t = 0+v(0+) = K1e(-1 + i1.732)0+ + K2e(-1 – i1.732)0+

1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2

Differentiating (i) with respect to ‘t’dv(t) = (-1 + i1.732)K1e(-1 + i1.732)t + (-1 – i1.732)K2e(-1 – i1.732)t

dtdv(0+) = (-1 + i1.732)K1e(-1 + i1.732)0+ + (-1 – i1.732)K2e(-1 – i1.732)0+

dt-1 = (-1 + i1.732)K1e0 + (-1 – i1.732)K2e0

-1 = (-1 + i1.732)K1(1) + (-1 – i1.732)K2(1)-1 = (-1 + i1.732)K1 + (-1 – i1.732)K2

(-1 + i1.732)K1 = (1 + i1.732)K2 - 1K1 = [0.5 – i0.866]K2 + 0.25 + i0.433 1 = [[0.5 – i0.866]K2 + 0.25 + i0.433] + K2

1 = [0.5 – i0.866]K2 + 0.25 + i0.433 + K2

1 - 0.25 - i0.433 = [0.5 – i0.866]K2 + K2

0.75 - i0.433 = [0.5 – i0.866 + 1]K2

0.75 - i0.433 = [1.5 – i0.866]K2

K2 = 0.51 = K1 + 0.5K1 = 0.5Putting corresponding values

v(t) = 0.5e(-1 + i1.732)t + 0.5e(-1 – i1.732)t

(c)d2v dv + 4 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 4sv + 2v = 0s2 + 4s + 2 = 0

a 1b 4c 2

-b b2 – 4acs1, s2 = 2a



-4 42 – 4(1)(2)s1, s2 = 2(1) -4 16 – 8s1, s2 = 2(1) -4 8s1, s2 = 2(1) -4 2.828s1, s2 = 2s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2s1, s2 = (-0.586), (-3.414)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-0.586)t + K2e(-3.414)t … (i)

At t = 0+v(0+) = K1e(-0.586)0+ + K2e(-3.414)0+

1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’dv(t) = (-0.586)K1e(-0.586)t + (-3.414)K2e(-3.414)t

dtdv(0+) = (-0.586)K1e(-0.586)0+ + (-3.414)K2e(-3.414)0+

dt-1 = (-0.586)K1e0 + (-3.414)K2e0

-1 = (-0.586)K1(1) + (-3.414)K2(1)-1 = -0.586K1 - 3.414K2

K1 = 1.706 – 5.826K2

1 = 1.706 – 5.826K2 + K2

0.706 = 4.826K2

K2 = 0.146K1 = 1.706 – 5.826(0.146)K1 = 1.706 – 0.851K1 = 0.855Putting corresponding values we get


v(t) = 0.855e(-0.586)t + 0.146e(-3.414)t

(d) d2v dv2 + 8 + 16v = 0 dt2 dtAfter simplificationd2v dv + 4 + 8v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 4sv + 8v = 0s2 + 4s + 8 = 0

a 1b 4c 8


-4 42 – 4(1)(8)s1, s2 = 2(1)

-4 16 – 32s1, s2 = 2(1) -4 -16s1, s2 = 2(1) -4 i4s1, s2 = 2s1, s2 = (-4 + i4)/2, (-4 – i4)/2s1, s2 = (-2 + i2), (-2 – i2)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t … (i)


At t = 0+v(0+) = K1e(-2 + i2)0+ + K2e(-2 – i2)0+

1 = K1e0 + K2e0


dtdv(0+) = (-2 + i2)K1e(-2 + i2)0+ + (-2 – i2)K2e(-2 – i2)0+ dt-1 = (-2 + i2)K1e0 + (-2 – i2)K2e0 -1 = (-2 + i2)K1(1) + (-2 – i2)K2(1)-1 = (-2 + i2)K1 + (-2 – i2)K2

K1 = -1i + 0.25 + i0.25K1 = -0.75i + 0.25 1 = [-0.75i + 0.25] + K2

1 + 0.75i – 0.25 = K2

0.75 + 0.75i = K2

v(t) = [-0.75i + 0.25]e(-2 + i2)t + [0.75 + 0.75i]e(-2 – i2)t

(e)d2v dv + 2 + 3v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 3v = 0s2 + 2s + 3 = 0

a 1b 2c 3


-2 22 – 4(1)(3)s1, s2 = 2(1) -2 4 – 12s1, s2 = 2(1)




v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t

At t = 0+v(0+) = K1e(-1 + i1.414)0+ + K2e(-1 – i1.414)0+

1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2

Differentiating (i) with respect to ‘t’dv(t) = (-1 + i1.414)K1e(-1 + i1.414)t + (-1 – i1.414)K2e(-1 – i1.414)t

dtdv(0+) = (-1 + i1.414)K1e(-1 + i1.414)0+ + (-1 – i1.414)K2e(-1 – i1.414)0+

dtdv(0+) = (-1 + i1.414)K1e0 + (-1 – i1.414)K2e0

dtdv(0+) = (-1 + i1.414)K1(1) + (-1 – i1.414)K2(1) dt-1 = (-1 + i1.414)K1 + (-1 – i1.414)K2

K1 = [0.334 – i0.943]K2 + 0.334 + i0.4711 = [0.334 – i0.943]K2 + 0.334 + i0.471 + K2

1 - 0.334 - i0.471 = [1.334 – i0.943]K2

(0.666 - i0.471) = [1.334 – i0.943]K2

K2 = 0.499 1 = K1 + 0.4990.501 = K1 Putting corresponding values we get

v(t) = 0.501e(-1 + i1.414)t + 0.499e(-1 – i1.414)t

(f)d2v dv


+ 3 + 5v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 3sv + 5v = 0s2 + 3s + 5 = 0

a 1b 3c 5




v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t … (i)

At t = 0+v(0+) = K1e(-1.5 + i1.659)0+ + K2e(-1.5 – i1.659)0+

1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’dv(t) = (-1.5 + i1.659)K1e(-1.5 + i1.659)t + (-1.5 – i1.659)K2e(-1.5 – i1.659)t

dtdv(0+)


= (-1.5 + i1.659)K1e(-1.5 + i1.659)0+ + (-1.5 – i1.659)K2e(-1.5 – i1.659)0+

dtdv(0+) = (-1.5 + i1.659)K1e0 + (-1.5 – i1.659)K2e0

dtdv(0+) = (-1.5 + i1.659)K1(1) + (-1.5 – i1.659)K2(1) dt-1 = (-1.5 + i1.659)K1 + (-1.5 – i1.659)K2

K1 = 0.401 – i0.6641 = 0.401 – i0.664 + K2

K2 = 0.599 + i0.664Putting corresponding values we get

v(t) = (0.401 – i0.664)e(-1.5 + i1.659)t + (0.599 + i0.664)e(-1.5 – i1.659)t

Q#6.8: Find particular solutions to the differential equations given in Prob. 6.4, given the initial conditions: dvv(0+) = 2, (0+) = 1 dt(a)d2v dv + 2 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 2sv + 2v = 0s2 + 2s + 2 = 0

a 1b 2c 2

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2 22 – 4(1)(2)s1, s2 = 2(1) -2 4 – 8s1, s2 = 2(1) -2 -4s1, s2 =


2(1) -2 i2s1, s2 = 2s1, s2 = (-2 + i2)/2, (-2 – i2)/2s1, s2 = (-1 + i1), (-1 – i1)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1)t + K2e(-1 – i1)t … (i)

At t = 0+v(0+) = K1e(-1 + i1)0+ + K2e(-1 – i1)0+

v(0+) = K1e0 + K2e0


dtdv(0+) = (-1 + i1)K1e(-1 + i1)0+ + (-1 – i1)K2e(-1 – i1)0+

dtdv(0+) = (-1 + i1)K1e0 + (-1 – i1)K2e0

dtdv(0+) = (-1 + i1)K1(1) + (-1 – i1)K2(1) dt1 = (-1 + i1)K1 + (-1 – i1)K2

K1 = -iK2 – 0.5 – i0.52 = -iK2 – 0.5 – i0.5 + K2

2 + 0.5 + i0.5 = -iK2 + K2

2.5 + i0.5 = [-i + 1]K2

K2 = 1 + i1.52 = K1 + 1 + i1.51 = K1 + i1.5K1 = 1 – i1.5

v(t) = (1 – i1.5)e(-1 + i1)t + (1 + i1.5)e(-1 – i1)t

(b)d2v dv + 2 + 4v = 0dt2 dt


Characteristic equation:as2 + bs + c = 0Heres2v + 2sv + 4v = 0s2 + 2s + 4 = 0

a 1b 2c 4


-2 22 – 4(1)(4)s1, s2 = 2(1) -2 4 – 16s1, s2 = 2(1)



v(t) = K1e(-1 + i1.732)t + K2e(-1 – i1.732)t … (i)

At t = 0+v(0+) = K1e(-1 + i1.732)0+ + K2e(-1 – i1.732)0+

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2

Differentiating (i) with respect to ‘t’dv(t)


= (-1 + i1.732)K1e(-1 + i1.732)t + (-1 – i1.732)K2e(-1 – i1.732)t

dtdv(0+) = (-1 + i1.732)K1e(-1 + i1.732)0+ + (-1 – i1.732)K2e(-1 – i1.732)0+

dt1 = (-1 + i1.732)K1e0 + (-1 – i1.732)K2e0

1 = (-1 + i1.732)K1(1) + (-1 – i1.732)K2(1)1 = (-1 + i1.732)K1 + (-1 – i1.732)K2

(-1 + i1.732)K1 = (1 + i1.732)K2 + 1K1 = [0.5 – i0.866]K2 - 0.25 - i0.433 2 = [0.5 – i0.866]K2 - 0.25 - i0.433 + K2

2 + 0.25 + i0.433 = [0.5 – i0.866]K2 + K2

2.25 + i0.433 = [0.5 – i0.866 + 1]K2

2.25 + i0.433 = [1.5 – i0.866]K2

K2 = 1 + i0.8672 = K1 + 1 + i0.8671 – i0.867 = K1 Putting corresponding values

v(t) = (1 – i0.867)e(-1 + i1.732)t + (1 + i0.867)e(-1 – i1.732)t

(c)

d2v dv + 4 + 2v = 0dt2 dtCharacteristic equation:as2 + bs + c = 0Heres2v + 4sv + 2v = 0s2 + 4s + 2 = 0

a 1b 4c 2


-4 42 – 4(1)(2)s1, s2 = 2(1) -4 16 – 8s1, s2 =


2(1) -4 8s1, s2 = 2(1) -4 2.828s1, s2 = 2s1, s2 = (-4 + 2.828)/2, (-4 – 2.828)/2s1, s2 = (-0.586), (-3.414)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-0.586)t + K2e(-3.414)t … (i)

At t = 0+v(0+) = K1e(-0.586)0+ + K2e(-3.414)0+

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’

dv(t) = (-0.586)K1e(-0.586)t + (-3.414)K2e(-3.414)t

dtdv(0+) = (-0.586)K1e(-0.586)0+ + (-3.414)K2e(-3.414)0+

dt1 = (-0.586)K1e0 + (-3.414)K2e0

1 = (-0.586)K1(1) + (-3.414)K2(1)1 = -0.586K1 - 3.414K2

K1 = -1.706 – 5.826K2 2 = [-1.706 – 5.826K2] + K2

2 = -1.706 – 5.826K2 + K2

2 + 1.706 = [–5.826 + 1]K2

3.706 = [–4.826]K2

K2 = -0.7682 = K1 + (-0.768)2.768 = K1 Putting corresponding values we get

v(t) = 2.768e(-0.586)t – 0.768e(-3.414)t

(d) d2v dv2 + 8 + 16v = 0


dt2 dtAfter simplification


a 1b 4c 8


-4 42 – 4(1)(8)s1, s2 = 2(1) -4 16 – 32s1, s2 = 2(1) -4 -16s1, s2 = 2(1) -4 i4s1, s2 = 2s1, s2 = (-4 + i4)/2, (-4 – i4)/2s1, s2 = (-2 + i2), (-2 – i2)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-2 + i2)t + K2e(-2 – i2)t

At t = 0+v(0+) = K1e(-2 + i2)0+ + K2e(-2 – i2)0+

2 = K1e0 + K2e0



dtdv(0+) = (-2 + i2)K1e(-2 + i2)0+ + (-2 – i2)K2e(-2 – i2)0+ dt1 = (-2 + i2)K1e0 + (-2 – i2)K2e0 1 = (-2 + i2)K1(1) + (-2 – i2)K2(1)1 = (-2 + i2)K1 + (-2 – i2)K2

K1 = -i – 0.25 – i0.25 K1 = -i1.25 – 0.25 2 = -i1.25 – 0.25 + K2

2 + i1.25 + 0.25 = K2

2.25 + i1.25 = K2

v(t) = [-i1.25 – 0.25]e(-2 + i2)t + [2.25 + i1.25]e(-2 – i2)t

(e)


a 1b 2c 3


-2 22 – 4(1)(3)s1, s2 = 2(1) -2 4 – 12s1, s2 = 2(1) -2 -8


s1, s2 = 2(1) -2 i2.828s1, s2 = 2s1, s2 = (-2 + i2.828)/2, (-2 – i2.828)/2s1, s2 = (-1 + i1.414), (-1 – i1.414)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1 + i1.414)t + K2e(-1 – i1.414)t

At t = 0+v(0+) = K1e(-1 + i1.414)0+ + K2e(-1 – i1.414)0+

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2

Differentiating (i) with respect to ‘t’

dv(t) = (-1 + i1.414)K1e(-1 + i1.414)t + (-1 – i1.414)K2e(-1 – i1.414)t

dtdv(0+) = (-1 + i1.414)K1e(-1 + i1.414)0+ + (-1 – i1.414)K2e(-1 – i1.414)0+

dtdv(0+) = (-1 + i1.414)K1e0 + (-1 – i1.414)K2e0

dtdv(0+) = (-1 + i1.414)K1(1) + (-1 – i1.414)K2(1) dt1 = (-1 + i1.414)K1 + (-1 – i1.414)K2

K1 = [0.334 – i0.943]K2 – 0.334 – i0.4712 = [0.334 – i0.943]K2 – 0.334 – i0.471 + K2

2 + 0.334 + i0.471 = [1.334 – i0.943]K2

2.334 + i0.471 = [1.334 – i0.943]K2

K2 = 1 + i1.0602 = K1 + 1 + i1.0601 - i1.060 = K1 Putting corresponding values we get

v(t) = (1 - i1.060)e(-1 + i1.414)t + (1 + i1.060)e(-1 – i1.414)t

(f)



a 1b 3c 5


-3 32 – 4(1)(5)s1, s2 = 2(1)

-3 9 – 20s1, s2 = 2(1) -3 -11s1, s2 = 2(1) -3 i3.317s1, s2 = 2s1, s2 = (-3 + i3.317)/2, (-3 – i3.317)/2s1, s2 = (-1.5 + i1.659), (-1.5 – i1.659)General solution:v(t) = K1es1t + K2es2t


v(t) = K1e(-1.5 + i1.659)t + K2e(-1.5 – i1.659)t … (i)

At t = 0+v(0+) = K1e(-1.5 + i1.659)0+ + K2e(-1.5 – i1.659)0+

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2 … (ii)Differentiating (i) with respect to ‘t’dv(t)


= (-1.5 + i1.659)K1e(-1.5 + i1.659)t + (-1.5 – i1.659)K2e(-1.5 – i1.659)t

dtdv(0+) = (-1.5 + i1.659)K1e(-1.5 + i1.659)0+ + (-1.5 – i1.659)K2e(-1.5 – i1.659)0+

dtdv(0+) = (-1.5 + i1.659)K1e0 + (-1.5 – i1.659)K2e0

dtdv(0+) = (-1.5 + i1.659)K1(1) + (-1.5 – i1.659)K2(1) dt1 = (-1.5 + i1.659)K1 + (-1.5 – i1.659)K2

K1 = 0.101 – i0.996 – 0.300 – i0.332K1 = -0.199 – i1.3282 = -0.199 – i1.328 + K2

2 + 0.199 + i1.328 = K2

2.199 + i1.328 = K2


v(t) = (-0.199 – i1.328)e(-1.5 + i1.659)t + (2.199 + i1.328)e(-1.5 – i1.659)t

Q#6.9: Solve the differential equation d3i d2i di3 + 8 + 10 + 3i = 0 dt dt dt3s3i + 8s2i + 10si + 3i = 0X-TICS EQUATION:3s3 + 8s2 + 10s + 3 = 0We can find roots using synthetic division

1 8 10 3

-1 -1 -07 -3

1 7 3 0 Remainder is zero.

s = -1So s + 1 is a factor.(s + 1)(s2 + 7s + 3) = 0s2 + 7s + 3 = 0

A 1B 7C 3

-b b2 – 4acs1, s2 =


2aPutting corresponding values we get -7 72 – 4(1)(3)s1, s2 = 2(1) -7 49 – 12s1, s2 = 2(1) -7 37s1, s2 = 2(1) -7 6.083s1, s2 = 2s1, s2 = (-7 + 6.083)/2, (-7 – 6.083)/2s1, s2 = (-0.917)/2, (-13.083)/2s1, s2 = -0.459, -6.542General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-0.459t + K2e-6.542t

So –1, -0.459, -6.542 are roots of the X-TICS equation.General solution:i(t) = K1e-1t + K2e-0.459t + K3e-6.542t

Q#6.10: Solve the differential equation d3i d2i di2 + 9 + 13 + 6i = 0 dt dt dt2s3i + 9s2i + 13si + 6i = 0X-TICS EQUATION:2s3 + 9s2 + 13s + 6 = 0We can find roots using synthetic division

2 9 13 6

-1 -2 -07 -6


s = -1So s + 1 is a factor.(s + 1)(2s2 + 7s + 6) = 02s2 + 7s + 6 = 0


A 2B 7C 6

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -7 72 – 4(1)(6)s1, s2 = 2(1) -7 49 – 24s1, s2 = 2(1) -7 25s1, s2 = 2(1)

-7 5s1, s2 = 2s1, s2 = (-7 + 5)/2, (-7 – 5)/2s1, s2 = (-2)/2, (-12)/2s1, s2 = -1, -6General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-1t + K2e-6t

So –1, -1, -6 are roots of the X-TICS equation.General solution:i(t) = K1e-1t + K2e-1t + K3e-6t … (i)At t = 0+i(0+) = K1e-1(0+) + K2e-1(0+) + K3e-6(0+)

0 = K1e0 + K2e0 + K3e0

0 = K1(1) + K2(1) + K3(1)

0 = K1 + K2 + K3

Differentiating (i) with respect to ‘t’di = -K1e-t - K2e-1t - 6K3e-6t

dtdi(0+) = -K1e-(0+) - K2e-1(0+) - 6K3e-6(0+)


dt1 = -K1e0 - K2e0 - 6K3e0

1 = -K1(1) - K2(1) - 6K3(1)

1 = -K1 - K2 - 6K3

d2i = K1e-t + K2e-1t + 6K3e-6t

dt2

d2i(0+) = K1e-(0+) + K2e-1(0+) + 6K3e-6(0+)

dt2

-1 = K1e0 + K2e0 + 6K3e0

-1 = K1(1) + K2(1) + 6K3(1)

-1 = K1 + K2 + 6K3

Q#6.11: The response of a network is found to be i = K1te-t, t 0where is real and positive. Find the time at which i(t) attains a maximum value.

= 4 K = 1t = 0 i(0) = 0t = 1 i(1) = 0.018t = 2 i(2) = 0.001t = 3 i(3) = 0



Q#6.12: In a certain network, it is found that the current is given by the expression i = K1e-1t – K2e-2t, t > 0, 1 > 2

Show that i(t) reaches a maximum value at time 1 1K1

t = ln 1 - 2 2K2

1 1K1

t = ln 1 K2

Solution:i(t) = K1e-1t – K2e-2t

t = 1 i(1) = K1e-1(1) – K2e-2(1)

i(1) = K1e-1 – K2e-2

t = 2 i(2) = K1e-1(2) – K2e-2(2)

i(2) = K1e-21 – K2e-22

t = 3 i(3) = K1e-1(3) – K2e-2(3)

i(3) = K1e-31 – K2e-32

t = (1/1)ln(1K1/K2)i((1/1)ln(1K1/K2)) = K1e-1((1/1)ln(1K1/K2))

– K2e-2((1/1)ln(1K1/K2))

1 1K1

t = ln 1 K2

Q#6.13: The graph shows a damped sinusoidal waveform having the general formKe-t sin(t + )From the graph, determine numerical values for K, , , and .t = 1 sec 1f = tf = 1 Hz = 2f = 2(1) = 2 = 6.28 rad/sec

t V(t)0 0

0.4 00.8 0

Q#6.14: Repeat Prob. 6.13 for the waveform of the accompanying figure.t = 4 msec 1f = tf = 250 Hz


= 2f = 2(250) = 1570 rad/sec = 00

Q#6.15: In the network of the figure, the switch K is closed and a steady state is reached in the network. At t = 0, the switch is opened. Find an expression for the current in the inductor, i2(t).Solution:Circuit diagram:

K 10 i2

+

100 V 1 H- 20 F

Equivalent circuit before switching:

At t = 0- 100i2(0-) = 10 i2(0-) = 10 A = i2(0+)VC(0-) = VC(0+) = 0Equivalent circuit after switching:


i2

Using KVL we have di2 1L + i2dt = 0 … (i) dt CDifferentiating we have d2i2 i2

L + = 0 dt2 CThe characteristic equation is i2

Ls2i2 + = 0 C 1Ls2 + = 0 C 1s2 = - LC

-1s = LC

1s = j LC

Roots are pure imaginary so from table 6-1Form of solutioni = K1cos t + K2 sin tHere

1


= LC

i = i2(t)i2(t) = K1cos (1/LC) 1/2 t + K 2 sin (1/LC) 1/2 t … (ii) From equation (i) at t = 0+ di2(0+)

L + VC(0+) = 0 dt Putting corresponding values

di2(0+)

= 0 Amp./sec dt

From (ii) at t = 0+i2(0+) = K1cos (1/LC)1/2(0+) + K2 sin (1/LC)1/2(0+) i2(0+) = K1(1) + 0K1 = 10 di2(t)

= -(1/LC)1/2K1sin (1/LC)1/2t + (1/LC)1/2K2 cos (1/LC)1/2t dtAt t = 0+di2(0+)

= -(1/LC)1/2K1sin (1/LC)1/2(0+) + (1/LC)1/2K2 cos (1/LC)1/2(0+) dtdi2(0+)

= -(1/LC)1/2K1sin (1/LC)1/2(0+) + (1/LC)1/2K2 cos (1/LC)1/2(0+) dt0 = (1/LC)1/2K2

0 = K2

i2(t) = (10)cos (1/LC)1/2(t) + (0) sin (1/LC)1/2(t) i2(t) = (10)cos (1/(1)(20(10-6)))1/2(t)

i2(t) = 10cos 223.607t

Q#6.16: The capacitor of the figure has an initial voltage VC(0-) = V1, and at the same time the current in the inductor is zero. At t = 0, the switch K is closed. Determine an expression for the voltage V2(t).Solution:


V1 V2

C R L

HereVC(0-) = VC(0+) = 0 = V2(0+)iL(0-) = iL(0+) = 0After switching dV2 V2 1C + + V2dt = 0 … (i) dt R LDifferentiating with respect to ‘t’ d2V2 dV2 V2

C + + = 0 dt2 Rdt Ld2V2 dV2 V2

+ + = 0 dt2 RCdt LCThe characteristic equation is s V2

s2V2 + V2 + = 0 RC LC s 1s2 + + = 0 RC LC

a 1b 1/RCc 1/LC


-(1/RC) (1/RC)2 – 4(1)(1/LC)s1, s2 = 2(1)

-(1/RC) (1/R2C2) – (4/LC)s1, s2 =


2s1, s2 = -(1/2RC) (1/2RC)2 – (1/LC) Case (a) if(1/2RC)2 – (1/LC) = 0(1/2RC)2 = (1/LC) that is 1 LR = 2 C Then 1s = - 2RCSo form of solution is:Nature of rootss1, s2 = (-1/2RC + 0), (-1/2RC - 0)s1, s2 = (-1/2RC), (-1/2RC)Negative real and equalV2(t) = K1es1t + K2tes2t

V2(t) = K1e(-1/2RC)t + K2te(-1/2RC)t … (ii)From equation (i) at t = 0+ dV2(0+) V2(0+)

C + + iL(0+) = 0 dt R Putting corresponding valuesdV2(0+) -V1

= dt RCAt t = 0+ V2(0+) = K1e(-1/2RC)0+ + K2(0+)e(-1/2RC)0+ … (ii)V1 = K1e0 V1 = K1

Differentiating (ii) with respect to ‘t’dV2(t) = (-1/2RC)K1e(-1/2RC)t + K2[-te(-1/2RC)t - e(-1/2RC)t] dtAt t = 0+dV2(0+) = (-1/2RC)K1e(-1/2RC)0+ + K2[-(0+)e(-1/2RC)0+ - e(-1/2RC)0+] dt

dV2(0+) = (-1/2RC)K1e0 + K2[-e0] dt


-V1

= (-1/2RC)K1(1) + K2[-(1)]RC V1

= (1/2RC)K1 + K2

RC Putting corresponding values V1

= (1/2RC)V1 + K2

RC -V1

K2 = 2RCPutting corresponding values

V1

V2(t) = V1e(-1/2RC)t - te(-1/2RC)t

2RC

Q#6.17: The voltage source in the network of the figure is described by the equation, V1 = 2 cos 2t for t 0 and is a short circuit prior to that time. Determine V2(t). Repeat if V1 = K1t for t 0 and V1 = 0 for t < 0.Solution:

V1 V2

Here iL(0-) = iL(0+) = 0 AVC(0-) = VC(0+) = 0 = V2(0+)For t 0 According to KCLV2 1 dV2 1 + + (V2 – V1)dt = 0 … (i) 2 2 dt 1Differentiating we have

dV2 1 d2V2

+ + V2 – V1 = 02dt 2 dt2 dV2 d2V2

+ -


+ + 2V2 – 2V1 = 0 dt dt2

dV2 d2V2

+ + 2V2 = 2V1 dt dt2

dV2 d2V2

+ + 2V2 = 2[2 cos 2t] dt dt2

dV2 d2V2

+ + 2V2 = 4 cos 2t … (ii) dt dt2

Equation (ii) is a non-homogeneous differential equationCharacteristic equation iss2V2 + sV2 + 2V2 = 0s2 + s + 2 = 0

a 1b 1c 2




V2(t) = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t … (ii)

Complementary solution.The form of particular solution is


V2P = Acos 2t + Bsin 2t … (iii)Substituting (iii) in (ii) d2 d [Acos 2t + Bsin 2t] + [Acos 2t + Bsin 2t] + 2[Acos 2t + Bsin 2t] = 4 cos 2t dt2 dtAfter simplification -2Asin 2t + 2B cos 2t – 4Acos 2t – 4Bsin 2t + 2Acos 2t + 2Bsin 2t = 4 cos 2tEquating coefficients of L.H.S & R.H.S-2A – 4B + 2B = 0-2A – 2B = 0-2A = 2B -A = B2B – 4A + 2A = 42B – 2A = 4B – A = 2-A – A = 2-2A = 2A = -1-(-1) = BB = 1V2P = -1cos 2t + 1sin 2tV2P = -cos 2t + sin 2tTherefore particular solution is V2 = complementary + particularV2 = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t + [-cos 2t + sin 2t]V2 = K1e(-0.5 + i1.323)t + K2e(-0.5 – i1.323)t - cos 2t + sin 2t … (iV)dV2

= (-0.5 + i1.323)K1e(-0.5 + i1.323)t + (-0.5 – i1.323)K2e(-0.5 – i1.323)t + 2sin 2t + 2cos 2t dt V2(0+) 1 dV2(0+)

+ + iL(0+) = 0 2 2 dt Putting corresponding valuesdV2(0+) = 0 dtAt t = 0+dV2(0+) = dt(-0.5 + i1.323)K1e(-0.5 + i1.323)0+ + (-0.5 – i1.323)K2e(-0.5 – i1.323)0+ + 2sin 2(0+) + 2cos 2(0+)0 = (-0.5 + i1.323)K1e0 + (-0.5 – i1.323)K2e0 + 2sin 0 + 2cos 00 = (-0.5 + i1.323)K1(1) + (-0.5 – i1.323)K2(1) + 2(1)0 = (-0.5 + i1.323)K1 + (-0.5 – i1.323)K2 + 2 … (V)At t = 0+ equation (iV)V2(0+) = K1e(-0.5 + i1.323)0+ + K2e(-0.5 – i1.323)0+ - cos 2(0+) + sin 2(0+)


0 = K1e0 + K2e0 – 10 = K1(1) + K2(1) – 10 = K1 + K2

– 11 = K1 + K2

… (Vi)K1 = 1 – K2

Putting the value of K1 in (V)0 = (-0.5 + i1.323)(1 – K2) + (-0.5 – i1.323)K2 + 2 0 = (-0.5 + i1.323) – K2(-0.5 + i1.323) + (-0.5 – i1.323)K2 + 2 0 = -0.5 + i1.323 + 0.5K2 – i1.323K2 – 0.5K2 - i1.323K2 + 2-1.5 - i1.323 = [0.5 – i1.323 – 0.5 - i1.323]K2

-1.5 - i1.323 = [–i2.646]K2

1.5 + i1.323 = [i2.646]K2

K2 = 0.5 – i0.567

Putting the value of K2 in (Vi)1 = K1 + 0.5 – i0.5670.5 + i0.567 = K1

V2 = [0.5 + i0.567]e(-0.5 + i1.323)t + [0.5 – i0.567]e(-0.5 – i1.323)t - cos 2t + sin 2t

Part (b)

If V1 = KtdV2 d2V2

+ + 2V2 = 2V1 dt dt2

dV2 d2V2

+ + 2V2 = 2K1 … (a) dt dt2

Equation (ii) is a non-homogeneous differential equationCharacteristic equation iss2V2 + sV2 + 2V2 = 0s2 + s + 2 = 0

A 1B 1C 2


-1 12 – 4(1)(2)s1, s2 =


2(1) -1 1 – 8s1, s2 = 2(1) -1 -7s1, s2 = 2(1) -1 i2.646s1, s2 = 2s1, s2 = (-1 + i2.646)/2, (-1 – i2.646)/2s1, s2 = (-0.5 + i1.323), (-0.5 – i1.323)General solution:v(t) = C1es1t + C2es2t


V2(t) = C1e(-0.5 + i1.323)t + C2e(-0.5 – i1.323)t … (i)

Complementary solution.The form of particular solution isFrom table 6.2A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn

V2P = At + B … (ii)Substituting equation (ii) in (a) d2 d [At + B] + [At + B] + 2[At + B] = 2K1t dt2 dtAfter simplificationA + 2At + 2B = 2K1tEquating coefficientst:2A = 2K1

A = K1

Constants:A + 2B = 0K1 + 2B = 0B = -0.5K1

V2P = K1t - 0.5K1

ThereforeParticular solution = Complementary + particularV2 = C1e(-0.5 + i1.323)t + C2e(-0.5 – i1.323)t + K1t - 0.5K1 … (ii)DifferentiatingdV2

= (-0.5 + i1.323)C1e(-0.5 + i1.323)t + (-0.5 – i1.323)C2e(-0.5 – i1.323)t + K1


dtAt t = 0+ eq. (ii)V2(0+) = C1e(-0.5 + i1.323)0+ + C2e(-0.5 – i1.323)0+ + K1(0+) - 0.5K1

0 = C1e0 + C2e0 - 0.5K1

0 = C1(1) + C2(1) - 0.5K1

0 = C1 + C2 - 0.5K1 … (iii)dV2(0+) = (-0.5 + i1.323)C1e(-0.5 + i1.323)0+ + (-0.5 – i1.323)C2e(-0.5 – i1.323)0+ + K1

dt0 = (-0.5 + i1.323)C1e0 + (-0.5 – i1.323)C2e0 + K1

0 = (-0.5 + i1.323)C1(1) + (-0.5 – i1.323)C2(1) + K1

0 = (-0.5 + i1.323)C1 + (-0.5 – i1.323)C2 + K1 … (iV)

Putting the value of C1 from (iii) into (iV) 0 = (-0.5 + i1.323)[0.5K1 – C2] + (-0.5 – i1.323)C2

+ K1

After simplification C2 = 0.250 – i0.284K1

Putting the value of C2 in (iii)0 = C1 + 0.250 – i0.284K1 - 0.5K1

C1 = -0.250 + i0.284K1 + 0.5K1

Putting corresponding values eq. (ii) becomesV2 = [-0.250 + i0.284K1 + 0.5K1]e(-0.5 + i1.323)t + [0.250 – i0.284K1]e(-0.5 – i1.323)t +

K1t - 0.5K1

Q#6.18: Solve the following nonhomogeneous differential equations for t 0.(a) d2i di + 2 + i = 1dt2 dtNonhomogeneous differential equationi(t) = iC + iP

Complementary solutionCharacteristic equation iss2i + 2si + 1i = 0s2 + 2s + 1 = 0

A 1B 2C 1


-2 22 – 4(1)(1)


s1, s2 = 2(1) -2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 = 2s1, s2 = (-2 + 0)/2, (-2 – 0)/2s1, s2 = -1, -1General solution:i(t) = K1es1t + K2tes2t


i(t) = K1e(-1)t + K2te(-1)t

Particular solutionFrom table 6.2

V(a constant) AiP = A = 1

i(t) = K1e(-1)t + K2te(-1)t + 1

(b)d2i di + 3 + 2i = 5t …(c)dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 3C 2


-3 32 – 4(1)(2)s1, s2 = 2(1)


-3 9 – 8s1, s2 = 2(1) -3 1s1, s2 = 2(1) -3 1s1, s2 = 2s1, s2 = (-3 + 1)/2, (-3 – 1)/2s1, s2 = (-1), (-2)General solution:v(t) = K1es1t + K2es2t


i(t) = K1e(-1)t + K2e(-2)t … (i)

Particular solutionFrom table 6.2A1tn B0tn + B1tn- 1 + … + Bn – 1t + Bn

iP = At + B … (ii)Putting the value of iP from (ii) into eq. (c)d2 d [At + B] + 3 [At + B] + 2[At + B] = 5tdt2 dtAfter simplification3A + 2At + 2B = 5tEquating coefficients2A = 5A = 2.5Constants3A + 2B = 03(2.5) + 2B = 07.5 + 2B = 0B = -3.75iP = 2.5t – 3.75iP = -15/4 + 5t/2i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2

(c)d2i di + 3 + 2i = 10 sin 10t … (k)dt2 dtNonhomogeneous differential equationi(t) = iC + iP



A 1B 3C 2


-3 32 – 4(1)(2)s1, s2 = 2(1) -3 9 – 8s1, s2 = 2(1) -3 1s1, s2 = 2(1) -3 1s1, s2 = 2s1, s2 = (-3 + 1)/2, (-3 – 1)/2s1, s2 = (-1), (-2)General solution:v(t) = K1es1t + K2es2t


i(t) = K1e(-1)t + K2e(-2)t … (i)

Particular solutionFrom table 6.2iP = A sin 10t + B cos 10tPutting the value of iP in (k)d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt10 sin 10t-100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t-100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10tEquating coefficientsSin


-100A – 30B + 2A = 10-98A – 30B = 10Cos-100B + 30A + 2B = 0-98B + 30A = 0B = 0.306A-98A – 30(0.306A) = 10-98A – 9.18A = 10-107.18A = 10A = -0.094B = 0.306(-0.094)B = -0.029NowiP = -0.094 sin 10t + (-0.029) cos 10tiP = -0.094 sin 10t - 0.029 cos 10ti(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t]

i(t) = K1e(-1)t + K2e(-2)t - 0.094 sin 10t - 0.029 cos 10t

(d)d2q dq + 5 + 6q = te-t … (m)dt2 dtNonhomogeneous differential equationq(t) = qC + qP

Complementary solutionCharacteristic equation iss2q + 5sq + 6q = 0s2 + 5s + 6 = 0

A 1B 5C 6


-5 52 – 4(1)(6)s1, s2 = 2(1) -5 25 – 24s1, s2 = 2(1) -5 1


s1, s2 = 2(1) -5 1s1, s2 = 2s1, s2 = (-5 + 1)/2, (-5 - 1)/2s1, s2 = (-4)/2, (-6)/2s1, s2 = -2, -3General solution:q(t) = K1es1t + K2es2t


q(t) = K1e-2t + K2e-3t

Particular solutionqP = (At + B)te-t multiplied by ‘t’ because e-t is repeating in both iP & iC

Putting the value of iP in eq. md2 d [(At + B)te-t] + 5 [(At + B)te-t] + 6[(At + B)te-t] = te-t dt2 dtAt2e-t - 2Ate-t – 2Ate-t + 2Ae-t + Bte-t – Be-t – Be-t + 5[-At2e-t + 2Ate-t - Bte-t + Be-t]+ 6[At2e-t + Bte-t] = te-t

At2e-t - 2Ate-t – 2Ate-t + 2Ae-t - Bte-t – Be-t – Be-t - 5At2e-t + 10Ate-t - 5Bte-t + 5Be-t

+ 6At2e-t + 6Bte-t = te-t

Equating coefficientst2e-t:

A – 5A + 6A = 02A = 0A = 0te-t:-2A – 2A + 10A = 16B = 3B = 0.5e-t:2 – B – B + 5B = 02 + 3B = 0B = -0.667

When B = 0.5qP = (0t + (0.5))te-t

qP = 0.5te-t

When B = -0.667qP = (0t + (-0.667))te-t

qP = -0.667te-t

q(t) = K1e-2t + K2e-3t - 0.667te-t, K1e-2t + K2e-3t + 0.5te-t


(e)d2V dV + 5 + 6V = e-2t + 5e-3t … (n)dt2 dtNonhomogeneous differential equationV(t) = VC + VP

Complementary solutionCharacteristic equation iss2V + 5sV + 6V = 0s2 + 5s + 6 = 0

A 1B 5C 6

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -5 52 – 4(1)(6)s1, s2 = 2(1) -5 25 – 24s1, s2 = 2(1) -5 1s1, s2 = 2(1) -5 1s1, s2 = 2s1, s2 = (-5 + 1)/2, (-5 - 1)/2s1, s2 = (-4)/2, (-6)/2s1, s2 = -2, -3General solution:V(t) = K1es1t + K2es2t


V(t) = K1e-2t + K2e-3t

Particular solutionVP = Ce-2t + Ce-3t

Substituting the value of VP in eq. (n)d2 d [Ce-2t + Ce-3t] + 5 [Ce-2t + Ce-3t] + 6[Ce-2t + Ce-3t] = e-2t + 5e-3t

dt2 dt4Ce-2t + 9Ce-3t + 5[-2Ce-2t – 3Ce-3t] + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t


4Ce-2t + 9Ce-3t - 10Ce-2t – 15Ce-3t + 6Ce-2t + 6Ce-3t = e-2t + 5e-3t

Equating coefficientse-2t

4C – 10C + 6C = 14C – 10C + 6C = 1C = 0VP = (0)e-2t + (0)e-3t

VP = 0V(t) = K1e-2t + K2e-3t + 0V(t) = K1e -2t + K 2e -3t

Q#6.19: Solve the differential equations given in Prob. 6.18 subject to the following initial conditions: dxx(0+) = 1 & (0+) = -1 dtwhere x is the general dependent variable.(a) d2i di + 2 + i = 1dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 2C 1

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2 22 – 4(1)(1)s1, s2 = 2(1) -2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 =


2s1, s2 = (-2 + 0)/2, (-2 – 0)/2s1, s2 = -1, -1General solution:i(t) = K1es1t + K2tes2t


i(t) = K1e(-1)t + K2te(-1)t



i(t) = K1e(-1)t + K2te(-1)t + 1

At t = 0+i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ + 11 = K1e0 + 11 = K1(1) + 1K1 = 0di (0+) = -K1e-(0+) - K2(0+)e-(0+) + K2e-(0+)

dt-1 = -K1e0 + K2e0

-1 = -K1(1) + K2(1)-1 = -K1

+ K2

-1 = -(0) + K2

-1 = K2

i(t) = (0)e(-1)t + (-1)te(-1)t + 1

i(t) = - te-t + 1

(b)d2i di + 3 + 2i = 5t …(c)dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 3C 2

-b b2 – 4ac


s1, s2 = 2aPutting corresponding values we get



i(t) = K1e(-1)t + K2e(-2)t … (i)


iP = At + B … (ii)Putting the value of iP from (ii) into eq. (c)d2 d [At + B] + 3 [At + B] + 2[At + B] = 5tdt2 dtAfter simplification3A + 2At + 2B = 5tEquating coefficients2A = 5A = 2.5Constants3A + 2B = 03(2.5) + 2B = 07.5 + 2B = 0B = -3.75iP = 2.5t – 3.75iP = -15/4 + 5t/2i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2 … (b)At t = 0+


i(0+) = K1e(-1)0+ + K2e(-2)0+ - 15/4 + 5(0+)/2 1 = K1e0 + K2e0 - 15/4 1 = K1(1) + K2(1) - 15/41 = K1 + K2

- 15/41 + 15/4 = K1 + K2

4.75 = K1 + K2 … (A)Differentiating (b) at t = 0+di (0+) = -K1e(-1)0+ - 2K2e(-2)0+ + 5/2dt-1 = -K1e0 - 2K2e0 + 5/2-1 = -K1(1) - 2K2(1) + 5/2-1 = -K1 - 2K2 + 5/2-1 – 2.5 = -K1 - 2K2

-3.5 = -K1 - 2K2

From (A)K1 = 4.75 – K2

-3.5 = -[4.75 – K2] - 2K2

-3.5 = -4.75 + K2 - 2K2

-3.5 + 4.75 = -K2

1.25 = -K2

K2 = -1.254.75 = K1 + (-1.25)4.75 + 1.25= K1 K1 = 6Putting corresponding values of K1 & K2

i(t) = 6e(-1)t + (-1.25)e(-2)t - 15/4 + 5t/2

i(t) = 6e(-1)t - 1.25e(-2)t - 15/4 + 5t/2

(c)d2i di + 3 + 2i = 10 sin 10t … (k)dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 3C 2

-b b2 – 4acs1, s2 = 2a





i(t) = K1e(-1)t + K2e(-2)t … (i)

Particular solutionFrom table 6.2iP = A sin 10t + B cos 10tPutting the value of iP in (k)d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt10 sin 10t-100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t-100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10tEquating coefficientsSin-100A – 30B + 2A = 10-98A – 30B = 10Cos-100B + 30A + 2B = 0-98B + 30A = 0B = 0.306A-98A – 30(0.306A) = 10-98A – 9.18A = 10-107.18A = 10A = -0.094


B = 0.306(-0.094)B = -0.029NowiP = -0.094 sin 10t + (-0.029) cos 10tiP = -0.094 sin 10t - 0.029 cos 10ti(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t]


At t = 0+i(0+) = K1e(-1)0+ + K2e(-2)0+ - 0.094 sin 10(0+) - 0.029 cos 10(0+) … (c)1 = K1e0 + K2e0 - 0.094 sin 0 - 0.029 cos 01 = K1(1) + K2(1) - 0.029 (1)1 = K1 + K2 - 0.029 1.029 = K1 + K2 … (y)Differentiating eq. (c) with respect to ‘t’di (0+) = -K1e-(0+) - 2K2e(-2)0+ - 0.094(10) cos 10(0+) + 0.029(10) sin 10(0+)dt-1 = -K1e0 - 2K2e0 - 0.94 cos 0 + 0.029(10) sin 0 -1 = -K1(1) - 2K2(1) - 0.94 (1) -1 = -K1 - 2K2

- 0.940.06 = K1 + 2K2

From eq. (y) K1 = 1.029 – K2 0.06 = [1.029 – K2] + 2K2

0.06 = 1.029 – K2 + 2K2

-0.969 = K2

Putting the value of K2 in (y)1.029 = K1 + (-0.969) 1.029 + 0.969 = K1 1.998 = K1 i(t) = 1.998e(-1)t + (-0.969)e(-2)t - 0.094 sin 10t - 0.029 cos 10t

i(t) = 1.998e(-1)t - 0.969e(-2)t - 0.094 sin 10t - 0.029 cos 10t(d)d2q dq + 5 + 6q = te-t … (m)dt2 dtNonhomogeneous differential equationq(t) = qC + qP


A 1


B 5C 6

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -5 52 – 4(1)(6)s1, s2 = 2(1) -5 25 – 24s1, s2 = 2(1) -5 1s1, s2 = 2(1) -5 1s1, s2 = 2s1, s2 = (-5 + 1)/2, (-5 - 1)/2s1, s2 = (-4)/2, (-6)/2s1, s2 = -2, -3General solution:q(t) = K1es1t + K2es2t


q(t) = K1e-2t + K2e-3t






A – 5A + 6A = 02A = 0A = 0te-t:-2A – 2A + 10A = 16B = 3


B = 0.5e-t:2 – B – B + 5B = 02 + 3B = 0B = -0.667

When B = 0.5qP = (0t + (0.5))te-t

qP = 0.5te-t

When B = -0.667qP = (0t + (-0.667))te-t

qP = -0.667te-t


At t = 0+q(0+) = K1e-2(0+) + K2e-3(0+) - 0.667(0+)e-(0+), K1e-2(0+) + K2e-3(0+) + 0.5(0+)e-(0+)

1 = K1e0 + K2e0, K1e0 + K2e0

1 = K1(1) + K2(1), K1(1) + K2(1)

1 = K1 + K2, K1 + K2 … (y)q(t) = K1e-2t + K2e-3t - 0.667te-t

Differentiating with respect to ‘t’ at t = 0+dq (0+) = -2K1e-2(0+) -3K2e-3(0+) - 0.667[-(0+)e-(0+) + e-(0+)]dt-1 = -2K1e0 -3K2e0 - 0.667[e0]-1 = -2K1(1) -3K2(1) - 0.667[1]-1 = -2K1 -3K2

- 0.6670.333 = 2K1 + 3K2

From eq. (y) K1 = 1 – K2

0.333 = 2[1 – K2] + 3K2

0.333 = 2 – 2K2 + 3K2

0.333 = 2 + K2

K2 = 0.333 - 2K2 = -1.6671 = K1 + (-1.667)K1 = 2.667q(t) = 2.667e-2t + (-1.667)e-3t - 0.667te-t

q(t) = 2.667e-2t - 1.667e-3t - 0.667te-t


Complementary solutionCharacteristic equation iss2V + 5sV + 6V = 0


s2 + 5s + 6 = 0A 1B 5C 6



V(t) = K1e-2t + K2e-3t






4C – 10C + 6C = 14C – 10C + 6C = 1C = 0VP = (0)e-2t + (0)e-3t

VP = 0V(t) = K1e-2t + K2e-3t + 0


V(t) = K1e -2t + K 2e -3t … (k) At t = 0+V(0+) = K1e-2(0+) + K2e-3(0+)

1 = K1e0 + K2e0

1 = K1(1) + K2(1)1 = K1 + K2 … (A)Differentiating eq. (k) with respect to ‘t’dV (0+) = -2K1e-2(0+) - 3K2e-3(0+)

dt-1 = -2K1e0 - 3K2e0

-1 = -2K1(1) - 3K2(1)-1 = -2K1 - 3K2 … (B)K1 = 1 – K2

-1 = -2[1 – K2] - 3K2 … (B)-1 = -2 + 2K2 - 3K2

-1 = K2

1 = K1 + (-1) … (A)2 = K1 V(t) = 2e-2t + (-1)e-3t … (k) V(t) = 2e -2t - e -3t Q#6.20: Find the particular solutions to the differential equations of Prob. 6.18 for the following initial conditions: dxx(0+) = 2 & (0+) = -1 dtwhere x is the general dependent variable.(a) d2i di + 2 + i = 1dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 2C 1



-2 22 – 4(1)(1)s1, s2 = 2(1) -2 4 – 4s1, s2 = 2(1) -2 0s1, s2 = 2(1) -2 0s1, s2 = 2s1, s2 = (-2 + 0)/2, (-2 – 0)/2s1, s2 = -1, -1General solution:i(t) = K1es1t + K2tes2t


i(t) = K1e(-1)t + K2te(-1)t



i(t) = K1e(-1)t + K2te(-1)t + 1

At t = 0+i(0+) = K1e(-1)0+ + K2(0+)e(-1)0+ + 12 = K1e0 + 12 = K1(1) + 1K1 = 1di (0+) = -K1e-(0+) - K2(0+)e-(0+) + K2e-(0+)

dt-1 = -K1e0 + K2e0

-1 = -K1(1) + K2(1)-1 = -K1

+ K2

-1 = -(1) + K2

0 = K2

i(t) = (1)e(-1)t + (0)te(-1)t + 1

i(t) = e-t + 1

(b)d2i di + 3 + 2i = 5t …(c)


dt2 dtNonhomogeneous differential equationi(t) = iC + iP


A 1B 3C 2




i(t) = K1e(-1)t + K2e(-2)t … (i)


iP = At + B … (ii)Putting the value of iP from (ii) into eq. (c)d2 d [At + B] + 3 [At + B] + 2[At + B] = 5tdt2 dtAfter simplification3A + 2At + 2B = 5tEquating coefficients


2A = 5A = 2.5Constants3A + 2B = 03(2.5) + 2B = 07.5 + 2B = 0B = -3.75iP = 2.5t – 3.75iP = -15/4 + 5t/2i(t) = K1e(-1)t + K2e(-2)t - 15/4 + 5t/2 … (b)At t = 0+i(0+) = K1e(-1)0+ + K2e(-2)0+ - 15/4 + 5(0+)/2 2 = K1e0 + K2e0 - 15/4 2 = K1(1) + K2(1) - 15/42 = K1 + K2

- 15/42 + 15/4 = K1 + K2

5.75 = K1 + K2 … (A)Differentiating (b) at t = 0+di (0+) = -K1e(-1)0+ - 2K2e(-2)0+ + 5/2dt-1 = -K1e0 - 2K2e0 + 5/2-1 = -K1(1) - 2K2(1) + 5/2-1 = -K1 - 2K2 + 5/2-1 – 2.5 = -K1 - 2K2

-3.5 = -K1 - 2K2

From (A)K1 = 5.75 – K2

-3.5 = -[5.75 – K2] - 2K2

-3.5 = -5.75 + K2 - 2K2

-3.5 + 5.75 = -K2

2.25 = -K2

K2 = -2.255.75 = K1 + (-2.25)5.75 + 2.25= K1 K1 = 8Putting corresponding values of K1 & K2

i(t) = 8e(-1)t + (-2.25)e(-2)t - 15/4 + 5t/2

i(t) = 8e(-1)t - 2.25e(-2)t - 15/4 + 5t/2

(c)d2i di + 3 + 2i = 10 sin 10t … (k)dt2 dtNonhomogeneous differential equation


i(t) = iC + iP


A 1B 3C 2




i(t) = K1e(-1)t + K2e(-2)t … (i)

Particular solutionFrom table 6.2iP = A sin 10t + B cos 10tPutting the value of iP in (k)d2 d [A sin 10t + B cos 10t] + 3 [A sin 10t + B cos 10t] + 2[A sin 10t + B cos 10t] = dt2 dt10 sin 10t-100A sin 10t – 100B cos 10t + 3[10A cos 10t - 10B sin 10t] + 2A sin 10t + 2B cos 10t = 10 sin 10t-100A sin 10t – 100B cos 10t + 30A cos 10t - 30B sin 10t + 2A sin 10t + 2B cos 10t = 10 sin 10tEquating coefficients


Sin-100A – 30B + 2A = 10-98A – 30B = 10Cos-100B + 30A + 2B = 0-98B + 30A = 0B = 0.306A-98A – 30(0.306A) = 10-98A – 9.18A = 10-107.18A = 10A = -0.094B = 0.306(-0.094)B = -0.029NowiP = -0.094 sin 10t + (-0.029) cos 10tiP = -0.094 sin 10t - 0.029 cos 10ti(t) = K1e(-1)t + K2e(-2)t + [-0.094 sin 10t - 0.029 cos 10t]


At t = 0+i(0+) = K1e(-1)0+ + K2e(-2)0+ - 0.094 sin 10(0+) - 0.029 cos 10(0+) … (c)2 = K1e0 + K2e0 - 0.094 sin 0 - 0.029 cos 02 = K1(1) + K2(1) - 0.029 (1)2 = K1 + K2 - 0.029 2.029 = K1 + K2 … (y)Differentiating eq. (c) with respect to ‘t’di (0+) = -K1e-(0+) - 2K2e(-2)0+ - 0.094(10) cos 10(0+) + 0.029(10) sin 10(0+)dt-1 = -K1e0 - 2K2e0 - 0.94 cos 0 + 0.029(10) sin 0 -1 = -K1(1) - 2K2(1) - 0.94 (1) -1 = -K1 - 2K2

- 0.940.06 = K1 + 2K2

From eq. (y) K1 = 2.029 – K2 0.06 = [2.029 – K2] + 2K2

0.06 = 2.029 – K2 + 2K2

-1.969 = K2

Putting the value of K2 in (y)1.029 = K1 + (-1.969) 1.029 + 1.969 = K1 2.998 = K1 i(t) = 2.998e(-1)t + (-1.969)e(-2)t - 0.094 sin 10t - 0.029 cos 10t

i(t) = 2.998e(-1)t - 1.969e(-2)t - 0.094 sin 10t - 0.029 cos 10t


(d)d2q dq + 5 + 6q = te-t … (m)dt2 dtNonhomogeneous differential equationq(t) = qC + qP


A 1B 5C 6


-5 52 – 4(1)(6)s1, s2 = 2(1) -5 25 – 24s1, s2 = 2(1) -5 1s1, s2 = 2(1)

-5 1s1, s2 = 2s1, s2 = (-5 + 1)/2, (-5 - 1)/2s1, s2 = (-4)/2, (-6)/2s1, s2 = -2, -3General solution:q(t) = K1es1t + K2es2t


q(t) = K1e-2t + K2e-3t







A – 5A + 6A = 02A = 0A = 0te-t:-2A – 2A + 10A = 16B = 3B = 0.5e-t:2 – B – B + 5B = 02 + 3B = 0B = -0.667

When B = 0.5qP = (0t + (0.5))te-t

qP = 0.5te-t

When B = -0.667qP = (0t + (-0.667))te-t

qP = -0.667te-t


At t = 0+q(0+) = K1e-2(0+) + K2e-3(0+) - 0.667(0+)e-(0+), K1e-2(0+) + K2e-3(0+) + 0.5(0+)e-(0+)

2 = K1e0 + K2e0, K1e0 + K2e0

2 = K1(1) + K2(1), K1(1) + K2(1)

2 = K1 + K2, K1 + K2 … (y)q(t) = K1e-2t + K2e-3t - 0.667te-t

Differentiating with respect to ‘t’ at t = 0+dq (0+) = -2K1e-2(0+) -3K2e-3(0+) - 0.667[-(0+)e-(0+) + e-(0+)]dt-1 = -2K1e0 -3K2e0 - 0.667[e0]-1 = -2K1(1) -3K2(1) - 0.667[1]-1 = -2K1 -3K2

- 0.6670.333 = 2K1 + 3K2

From eq. (y) K1 = 1 – K2

0.333 = 2[2 – K2] + 3K2

0.333 = 4 – 2K2 + 3K2


0.333 = 4 + K2

K2 = 0.333 - 4K2 = -3.6671 = K1 + (-3.667)K1 = 4.667q(t) = 4.667e-2t + (-3.667)e-3t - 0.667te-t

q(t) = 4.667e-2t - 3.667e-3t - 0.667te-t


Complementary solutionCharacteristic equation iss2V + 5sV + 6V = 0s2 + 5s + 6 = 0

a 1b 5c 6



V(t) = K1e-2t + K2e-3t







4C – 10C + 6C = 14C – 10C + 6C = 1C = 0VP = (0)e-2t + (0)e-3t

VP = 0V(t) = K1e-2t + K2e-3t + 0V(t) = K1e -2t + K 2e -3t … (k) At t = 0+V(0+) = K1e-2(0+) + K2e-3(0+)

2 = K1e0 + K2e0

2 = K1(1) + K2(1)2 = K1 + K2 … (A)Differentiating eq. (k) with respect to ‘t’dV (0+) = -2K1e-2(0+) - 3K2e-3(0+)

dt-1 = -2K1e0 - 3K2e0

-1 = -2K1(1) - 3K2(1)-1 = -2K1 - 3K2 … (B)K1 = 2 – K2

-1 = -2[2 – K2] - 3K2 … (B)-1 = -4 + 2K2 - 3K2

-3 = K2

1 = K1 + (-3) … (A)4 = K1 V(t) = 4e-2t + (-3)e-3t … (k) V(t) = 4e -2t - 3e -3t

Q#6.21: Solve the differential equation d3i d2i di2 + 9 + 13 + 6i = K0te-tsint … (u) dt dt dt2s3i + 9s2i + 13si + 6i = 0X-TICS EQUATION:2s3 + 9s2 + 13s + 6 = 0


We can find roots using synthetic division

2 9 13 6

-1 -2 -07 -6


s = -1So s + 1 is a factor.(s + 1)(2s2 + 7s + 6) = 02s2 + 7s + 6 = 0

A 2B 7C 6

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -7 72 – 4(1)(6)s1, s2 = 2(1) -7 49 – 24s1, s2 = 2(1) -7 25s1, s2 = 2(1) -7 5s1, s2 = 2s1, s2 = (-7 + 5)/2, (-7 – 5)/2s1, s2 = (-2)/2, (-12)/2s1, s2 = -1, -6General solution:i(t) = K1es1t + K2es2t


i(t) = K1e-1t + K2e-6t

So –1, -1, -6 are roots of the X-TICS equation.General solution:i(t) = K1e-1t + K2e-1t + K3e-6t … (i)Particular olutioniP = Fte-tcos t + Gte-tsin t


Putting the value of iP in (u)Here in eq. (u) i = iP

SimplificationI st derivative of i P:= F(1)e-tcos t + Ft(-e-t)cos t + Fte-t(-sin t) + G(1)e-tsin t + Gt(-e-t)sin t + Gte-tcos t

= Fe-tcos t - Fte-tcos t - Fte-tsin t + Ge-tsin t - Gte-tsin t + Gte-tcos t

2 nd derivative of i P:= F[-e-tsin t – e-tcos t] – F[e-tcos t - te-tcos t - te-tsin t] – F[e-tsin t - te-tsin t + te-tcos t] + G[e-tcos t – e-tsin t] – G[e-tsin t - te-tsin t + te-tcos t] + G[e-tcos t - te-tcos t - te-tsin t]

= -2Fe-tsin t – 2Fe-tcos t + 2Fte-tsin t + 2Ge-tcos t – 2Ge-tsin t - 2Gte-tcos t

3 rd derivative of i P:= -2F[e-tcos t – e-tsin t] - 2F[-e-tsin t – e-tcos t] + 2F[e-tsin t - te-tsin t + te-tcos t] + 2G[-e-tsin t – e-tcos t] – 2G[e-tcos t – e-tsin t] – 2G[e-tcos t - te-tcos t - te-tsin t]

= 6Fe-tsin t – 2Fte-tsin t + 2Fte-tcos t – 6Ge-tcos t + 2Gte-tcos t + 2Gte-tsin t

2[6Fe-tsin t – 2Fte-tsin t + 2Fte-tcos t – 6Ge-tcos t + 2Gte-tcos t + 2Gte-tsin t] +

9[-2Fe-tsin t – 2Fe-tcos t + 2Fte-tsin t + 2Ge-tcos t – 2Ge-tsin t - 2Gte-tcos t] +

13[Fe-tcos t - Fte-tcos t - Fte-tsin t + Ge-tsin t - Gte-tsin t + Gte-tcos t] +

6[Fte-tcos t + Gte-tsin t] = K0te-tsin t

12Fe-tsin t – 4Fte-tsin t + 4Fte-tcos t – 12Ge-tcos t + 4Gte-tcos t + 4Gte-tsin t

- 18Fe-tsin t – 4Fe-tcos t + 4Fte-tsin t + 4Ge-tcos t – 4Ge-tsin t - 4Gte-tcos t +

13Fe-tcos t - 13Fte-tcos t - 13Fte-tsin t + 13Ge-tsin t - 13Gte-tsin t + 13Gte-tcos t +

6Fte-tcos t + 6Gte-tsin t = K0te-tsin tComparing coefficientse-tsin t:12F – 18F – 4G + 13G = 0 -6F + 9G = 0 -2F + 3G = 0 … (7)te-tsin t:-4F + 4G + 4F – 13F – 13G + 6G = K0

-3G – 13F = K0 … (8)From eq. (7) F = 1.5G put in (8)-3G – 13(1.5G) = K0

-3G – 19.5G = K0

-22.5G = K0


G = -0.045K0

Put in (7)-2F + 3(-0.045K0) = 0-2F – 0.135K0 = 0F = -0.068K0

iP = (-0.068K0)te-tcos t + (-0.045K0)te-tsin tiP = -0.068K0te-tcos t - 0.045K0te-tsin ti(t) = K1e-1t + K2e-1t + K3e-6t + (-0.068K0te-tcos t - 0.045K0te-tsin t)

i(t) = K1e-1t + K2e-1t + K3e-6t - 0.068K0te-tcos t - 0.045K0te-tsin t … (9)

At t = 0+i(0+) = K1e-1(0+) + K2e-1(0+) + K3e-6(0+) - 0.068K0(0+)e-(0+)cos (0+) - 0.045K0(0+)e-(0+)sin (0+)1 = K1e0 + K2e0 + K3e0

1 = K1(1) + K2(1) + K3(1)

1 = K1 + K2 + K3

Differentiating eq. (9) with respect to ‘t’di (0+) = -K1e-(0+) - K2e-1(0+) - 6K3e-6(0+) - 0.068K0[e-(0+)cos (0+) dt– (0+)e-(0+)cos (0+) – (0+)e-(0+)sin (0+)] - 0.045K0[e-(0+)sin (0+) – (0+)e-(0+)sin (0+) + (0+)e-

(0+)cos (0+)]After simplification-1 = -K1 – K2 – 6K3 - 0.068K0[1] -1 = -K1 – K2 – 6K3 - 0.068K0

1 = K1 + K2 + 6K3 + 0.068K0

Q#6.22: A special generator has a voltage variation given by the equation V(t) = t V, where t is the time in seconds and t 0. This generator is connected to an RL series circuit, where R = 2 and L = 1 H, at time t = 0 by the closing of a switch. Find the equation for the current as a function of time i(t).

K L

V(t) iR

At t 0According to KVL

+ -


diL + iR = V(t) … (c) dtPutting corresponding valuesdi + 2i = tdtNonhomogeneous differential equationV(t) = VC + VP

Complementary solutionCharacteristic equation issi + 2i = 0s + 2 = 0s = -2iC = Kest

iC = Ke(-2)t

iC = Ke-2t


iP = At + B … (ii)Putting the value of iP from (ii) into eq. (c)d [At + B] + 2[At + B] = tdtAfter simplificationsA + 2At + 2B = tEquating coefficientst:2A = 1A = 0.5Constants:A + 2B = 00.5 + 2B = 0B = -0.25iP = 0.5t + (-0.25) … (ii)iP = 0.5t - 0.25

i(t) = Ke-2t + 0.5t - 0.25

At t = 0+i(0+) = Ke-2(0+) + 0.5(0+) - 0.250 = Ke0 - 0.25 i(0+) = i(0-) = 0 (due to inductor)0 = K(1) - 0.25K = 0.25i(t) = 0.25e-2t + 0.5t - 0.25


Q#6.23: A bolt (flash of lighting) of lightning having a waveform which is approximated as V(t) = te-t strikes a transmission line having resistance R = 0.1 and inductance L = 0.1 H. Find i(t) =?Solution:

At t 0According to KVL diL + iR = V(t) dtPutting corresponding values di(0.1) + (0.1)i = te-t … (h) dtNonhomogeneous differential equationV(t) = VC + VP

Complementary solutionCharacteristic equation is0.1si + 0.1i = 0s + 1 = 0s = -1iC = Kest

iC = Ke(-1)t

iC = Ke-1t

Particular solutionFrom table 6.2Particular solutioniP = (At + B)te-t multiplied by ‘t’ because e-t is repeating in both iP & iC

Putting the value of qP in eq. (h)0.1[-At2e-t + 2Ate-t - Bte-t + Be-t] + 0.1[At2e-t + Bte-t] = te-t

-0.1At2e-t + 0.2Ate-t – 0.1Bte-t + 0.1Be-t + 0.1At2e-t + 0.1Bte-t] = te-t

Equating coefficients:t2e-t:-0.1A + 0.1A = 0te-t:0.2A – 0.1B + 0.1B = 10.2A = 1A = 5e-t:

+ -


0.1B = 0B = 0iP = (5t + 0)te-t

iP = 5t2e-t

i(t) = Ke-1t + 5t2e-t

i(t) = Ke-t + 5t2e-t

At t = 0+i(0+) = Ke-(0+) + 5(0+)2e-(0+)

0 = Ke0

0 = Ki(t) = 0e-1t + 5t2e-t

i(t) = 5t2e-t

i(0+) = i(0-) = 0 (due to inductor)

Q#6.24: In the network of the figure, the switch K is closed at t = 0 with the capacitor initially unenergized. For the numerical values given, find i(t).

HereVC(0-) = VC(0+) = 0 V& 10e-(0+) sin (0+) u(0+)i(0+) = 10 10e0 sin (0) u(0+)i(0+) = 10i(0+) = 0 Amp.For t 0, KVL 110i + idt = 10e-t sin t u(t) (10(10-6)) 110i + idt = 10e-t sin t (10(10-6))After simplificationi + 104idt = e-t sin t

+ -


Differentiating with respect to ‘t’di + 104i = e-t (cost - sin t) … (1)dtIt is a nonhomogeneous differential equationCharacteristic equation issi + 104i = 0s + 104 = 0s = -104

Therefore complementary function isiC = Ke-10, 000t

The form of particular integral isFrom table 6.2iP = (Acost + Bsint)e-t

Putting the value of iP in eq. (1)After simplification -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000[(Acost + Bsint)e-t] = e-t (cost - sin t)-Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000Acoste-t + 10, 000Bsinte-t = e-t cost - e-t sin tEquating coefficientse-tsint:-A – B + 10, 000B = -1-A + 9999B = -1 … (2)e-tcost:-A + B + 10, 000A = 1B + 9999A = 1 … (3)From (2)A = 1 + 9999BB + 9999[1 + 9999B] = 1 … (3)B + 9999 + 9999(9999)B = 1B + 99980001B = -9998B[1 + 99980001] = -9998B[99980002] = -9998B = -10-4

A = 1 + 9999(-1/10000)A = 1 – 0.9999A = 10-4

iP = (10-4cost + (-10-4)sint)e-t

iP = (10 -4 cost - 10 -4 sint)e -t Complete solution isi(t) = Ke-10, 000t + (10-4cost - 10-4sint)e-t

At t = 0+i(0+) = Ke-10, 000(0+) + (10-4cos(0+) - 10-4sin(0+))e-(0+)

0 = Ke0 + (10-4(1) - 10-4(0))e0

0 = K(1) + (10-4(1) - 10-4(0))(1)0 = K + 10-4


K = -10-4 i(t) = (-10 -4 )e- 10, 000t + (10 -4 cost - 10 -4 sint)e -t

Q#6.25: In the network shown in the accompanying figure, a steady state is reached with the switch K open. At t = 0, the switch is closed. For the element values given, determine the current, i(t) for t 0. Solution:

0.5 FR = 1000

100 sin 377 tL = 1 H

Equivalent circuit before switching:By phasors:V = 10000

jXL = jLjXL = j(377)(1)jXL = j377 1-jXC = -j C-jXC = -j5305.04Therefore VI = R + jXL - jXC

After simplificationI = 0.019878.530

That is i(t) = 0.0198sin(377t + 78.530) AlsoVC = I(-jXC)VC = (0.019878.530)(5305.04-900)VC = 105.040-11.470

ThereforeVC(t) = 105.040sin(377t + (-11.470))

VC(t) = 105.040sin(377t - 11.470)

At t = 0+i(0+) = 0.0198sin(377(0+) + 78.530)i(0+) = 0.0198sin(78.530)i(0+) = 0.0198[0.981]i(0+) = 0.019 ampAt t = 0+

+ -


VC(0+) = 105.040sin(377(0+) - 11.470)VC(0+) = 105.040sin(-11.470)VC(0+) = 105.040[-sin(11.470)]VC(0+) = 105.040[-0.199]VC(0+) = -20.903 VoltsFor t 0 According to KVL di 1L + idt = 100sin377t … (d) dt CDifferentiating with respect to ‘t’ d2i iL + = 100(377)cos377t dt2 Cd2i i 37700cos377t + = dt2 LC Ld2i i 37700cos377t + = dt2 (1)(0.5(10-6)) 1d2i + 2(106)i = 37700cos377t … (5)dt2 s2i + 2(106)i = 37700cos377tComplementary solution:s2 + 2(106) = 0s = j1414.21iC = K1cos1414.21 t + K2sin1414.21 tParticular solution:iP = Acos377t + Bsin377tPutting the value of iP in eq. (5)After simplification-(377)2Acos377t – (377)2Bsin(377)t + 2(106)[Acos(377)t + Bsin(377)t] = 37700cos377t-(377)2Acos377t – (377)2Bsin377t + 2000000Acos377t + 2000000Bsin377t = 37700cos377tEquating coefficients:cos377t:-(377)2A + 2000000A = 37700-142129A + 2000000A = 377001857871A = 37700A = 0.020sin377t:–(377)2B + 2000000B = 0B = 0iP = 0.02cos377t + (0)sin377tiP = 0.02cos377t


Complete solution:i(t) = K1cos1414.21 t + K2sin1414.21 t + 0.02cos377t … (Q)At t = 0+i(0+) = K1cos1414.21(0+) + K2sin1414.21(0+) + 0.02cos377(0+)0.019 = K1cos(0) + 0.02cos(0)= K1(1) + 0.02(1)= K1 + 0.02K1 = 0.019 – 0.02K1 = -0.001Differentiating eq. (Q) with respect to ‘t’di(t) = -1414.21K1sin1414.21 t + 1414.21K2cos1414.21 t + 0.02(-377s)sin377t dtFrom differential eq. (d) at t = 0+di(0+) + VC(0+) = 100sin377t … (d) dt di(0+) + VC(0+) = 100sin377(0+) dt di(0+) + (-20.903) = 0 dt

di(0+) = 20.903 dt

20.903 = -1414.21K1sin1414.21(0+) + 1414.21K2cos1414.21(0+) + 0.02(-377s)sin377(0+)20.903 = -1414.21K1sin0 + 1414.21K2cos0 + 0.02(-377s)sin020.903 = 1414.21K2(1) + 020.903 = 1414.21K2

K2 = 0.015Therefore

i(t) = -0.001cos1414.21 t + 0.015sin1414.21 t + 0.02cos377t Q#6.26: In the network shown in figure, a steady state is reached with the switch K open. At t = 0, the value of the resistor R is changed to the critical value, Rcr defined by equation (6.88). For the element values given, determine the current i(t) for t 0.Solution:

LRcr = 2 C


Rcr = 2(1414.214)Rcr = 2828.428 For t 0 According to KVL di 1Rcri + L + idt = 100sin377t … (x) dt CDifferentiating with respect to ‘t’ di d2i iRcr + L + = 37700cos377t dt dt2 CAfter simplification di d2i2828.428 + (1) + 2(106)i = 37700cos377t … (w) dt dt2

Characteristic equation iss2i + 2828.428si + 2(106)i = 0s2 + 2828.428s + 2(106) = 0

A 1B 2828.428C 2(106)

-b b2 – 4acs1, s2 = 2aPutting corresponding values we get -2828.428 2828.4282 – 4(1)(2000000)s1, s2 = 2(1) -2828.428 8000004.952 – 8000000s1, s2 = 2(1) -2828.428 4.952s1, s2 = 2(1) -2828.428 2.225s1, s2 = 2(1)s1, s2 = (-2828.428 + 2.225)/2, (-2828.428 - 2.225)/2s1, s2 = -1413.102, -1415.326

i(t) = K1e-1413.102t + K2e-1415.326t


Form of particular integral isiP = Acos377t + Bsin377tPutting the value of iP in eq. (w)-(377)2Acos377t – (377)2Bsin377t + 2828.428[-377Asin377t + B377sin377t] + 2000000[Acos377t + Bsin377t] = 37700cos377t-142129Acos377t – 142129Bsin377t - 1066317.356Asin377t + 1066317.356Bsin377t+ 2000000Acos377t + 2000000Bsin377t = 37700cos377tEquating coefficients:cos377t:-142129A + 2000000A = 37700A = 0.020sin377t:–142129B - 1066317.356A + 1066317.356B + 2000000B = 02924188.356B - 1066317.356(0.020) = 02924188.356B = 1066317.356(0.020) 2924188.356B = 21326.34712B = 0.007iP = 0.020cos377t + 0.007sin377tComplete solution:

i(t) = K1e-1413.102t + K2e-1415.326t + 0.020cos377t + 0.007sin377t … (a)

At t = 0+i(0+) = K1e-1413.102(0+) + K2e-1415.326(0+) + 0.020cos377(0+) + 0.007sin377(0+)0.019 = K1e0 + K2e0 + 0.0200.019 = K1(1) + K2(1) + 0.020-0.001 = K1 + K2 … (z)Differentiating eq. (a) with respect to ‘t’ at t = 0+di (0+) = -1413.102K1e-1413.102(0+) - 1415.326K2e-1415.326(0+) + 0.020(-377)sin377(0+) + dt0.007(377)cos377(0+) … (g)At t = 0+ eq. (x) givesdi (0+) = -32.84dt-32.84 = -1413.102K1e-1413.102(0+) - 1415.326K2e-1415.326(0+) + 0.020(-377)sin377(0+) + 0.007(377)cos377(0+) … (g)-32.84 = -1413.102K1e0 - 1415.326K2e0 + 0.007(377)cos 0-32.84 = -1413.102K1(1) - 1415.326K2(1) + 0.007(377)(1)-32.84 = -1413.102K1 - 1415.326K2 + 2.639-32.84 – 2.639 = -1413.102K1 - 1415.326K2

-35.479 = -1413.102K1 - 1415.326K2

35.479 = 1413.102K1 + 1415.326K2

From eq. (z)K1 = -0.001 – K2

35.479 = 1413.102[-0.001 – K2] + 1415.326K2


35.479 = 1413.102[-0.001] – 1413.102K2 + 1415.326K2

35.479 = -1.413102 + 2.224K2

K2 = 16.588K1 = -0.001 – 16.588K1 = -16.589

i(t) = [-16.589]e-1413.102t + [16.588]e-1415.326t + 0.020cos377t + 0.007sin377t

Q#6.27: Consider the network shown in figure P6.24. The capacitor has an initial voltage, VC = 10 V. At t = 0, the switch K is closed. Determine i(t) for t 0.Solution:HereVC(0-) = VC(0+) = 10 V V(t) – VC(t)i(t) = 10 V(0+) – VC(0+)i(0+) = 10V(0+) = 10e-(0+) sin (0+) u(0+)V(0+) = 10(1)(0)(1)V(0+) = 0 Volts 0 – 10i(0+) = 10i(0+) = -1 amp.For t 0, KVL 110i + idt = 10e-t sin t u(t) (10(10-6)) 110i + idt = 10e-t sin t (10(10-6))After simplificationi + 104idt = e-t sin tDifferentiating with respect to ‘t’di + 104i = e-t (cost - sin t) … (1)dtIt is a nonhomogeneous differential equationCharacteristic equation issi + 104i = 0s + 104 = 0s = -104

Therefore complementary function isiC = Ke-10, 000t


The form of particular integral isFrom table 6.2iP = (Acost + Bsint)e-t

Putting the value of iP in eq. (1)After simplification -Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000[(Acost + Bsint)e-t] = e-t (cost - sin t)-Ae-tsint – Ae-tcost + Be-tcost - Be-tsint + 10, 000Acoste-t + 10, 000Bsinte-t = e-t cost - e-t sin tEquating coefficientse-tsint:-A – B + 10, 000B = -1-A + 9999B = -1 … (2)e-tcost:-A + B + 10, 000A = 1B + 9999A = 1 … (3)From (2)A = 1 + 9999BB + 9999[1 + 9999B] = 1 … (3)B + 9999 + 9999(9999)B = 1B + 99980001B = -9998B[1 + 99980001] = -9998B[99980002] = -9998B = -10-4

A = 1 + 9999(-1/10000)A = 1 – 0.9999A = 10-4

iP = (10-4cost + (-10-4)sint)e-t

iP = (10 -4 cost - 10 -4 sint)e -t Complete solution isi(t) = Ke-10, 000t + (10-4cost - 10-4sint)e-t

At t = 0+i(0+) = Ke-10, 000(0+) + (10-4cos(0+) - 10-4sin(0+))e-(0+)

-1 = Ke0 + (10-4(1) - 10-4(0))e0

-1 = K(1) + (10-4(1) - 10-4(0))(1)-1 = K + 10-4

K = -1 - 10-4 i(t) = (-1 - 10 -4 )e- 10, 000t + (10 -4 cost - 10 -4 sint)e -t Q#6.28: The network of the figure is operating in the steady state with the switch K open. At t = 0, the switch is closed. Find an expression for the voltage, V(t) for t 0.Solution:


Equivalent circuit before switching:

V(0-) = V(0+) = I0sin t [R1 + R2]

Equivalent circuit after switching:


For t 0According to KCL dV(t) V(t)C + = I0sin t dt R1

dV(t) V(t) I0sin t + = … (u) dt CR1 CIt is a nonhomogeneous differential equationCharacteristic equation is 1sV(t) + V(t) = 0 CR1

1s + = 0 CR1

1s = - CR1 Therefore complementary function isVC = Ke-(1/CR1)t

The form of particular integral isVP = Asin t + Bcos tPutting the value of VP in eq. (u)Acos t - Bsin t + [1/CR1][Asin t + Bcos t] = (I0sin t)/CAcos t - Bsin t + [1/CR1]Asin t + [1/CR1]Bcos t = (I0sin t)/CEquating coefficients:cos t:A + [1/CR1]B = 0 A = -B/CR1

sin t:-B + [1/CR1]A = I0/C-B + [1/CR1][-B/CR1] = I0/C

B = I0/C[- - 1/C2R12]

A = -[I0/C[- - 1/C2R12]]/CR1


Complete solutionV(t) = Ke-(1/CR1)t + [-[I0/C[- - 1/C2R1

2]]/CR1]sin t + [I0/C[- - 1/C2R12]]cos t

At t = 0+V(0+) = Ke-(1/CR1)0+ + [-[I0/C[- - 1/C2R1

2]]/CR1]sin (0+) + [I0/C[- - 1/C2R12]]cos

(0+)I0sin t [R1 + R2] = K(1) + [I0/C[- - 1/C2R1

2]]I0sin t [R1 + R2] = K + [I0/C[- - 1/C2R1

2]]

K = I0sin t [R1 + R2] - [I0/C[- - 1/C2R12]]

V(t) = [I0sin t [R1 + R2] - [I0/C[- - 1/C2R12]]]e-(1/CR1)t + [-[I0/C[- - 1/C2R1

2]]/CR1]sin t + [I0/C[- - 1/C2R1

2]]cos t

Q#6.29: Consider a series RLC network which is excited by a voltage source.Determine the characteristic equation.Locus of the roots of the equation.Plot the roots of the equation.Solution:

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Network analysis by van valkenburg solutions CH#6 part 1