Ordinary Differential

Equations (ODEs) “How to find solution of ODEs through

method of separating variables”

2014

Maria

Mahail Math World

7/24/2014

INTRODUCTION:- DIFFERENTIAL EQUATIONS:- An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called differential equations. For e.g.

i) )*)+ + y cos x = sin x

ii) )0*)+0 + xy 1)*

)+23 = 0

iii) 51 + 1)*)+237

80= )0*

)+0

are differential equations. ORDINARY DIFFERENTIAL EQUATIONS (ODEs):- A differential equation, in which ordinary derivatives of the dependent variable with respect to a single independent variable occur, is called “ordinary differential equation”, equations (i), (ii) and (iii) in afore-mentioned definitions are the examples of ODEs. ORDER OF DIFFERENTIAL EQUATIONS:- The order of the differential equation is the order of the highest derivative that occurs in the equation. Like in above equation (i) the order is 1 and that of the remaining two is 2. DEGREE OF DIFFERENTIAL EQUATIONS:- The degree of the differential equation is the degree of the highest order derivative that appears in the equation. Remember, the dependent variable and its derivative should be expressed in a form free from radicals and fractions. See the table below regarding the afore-mentioned.

Equation no. Order Degree i 1 1 ii 2 1 iii 2 2

SOLUTION OF DIFFERENTIAL EQUATIONS:- A solution or integral of a differential equation is a relation between variables, not containing derivatives, such that this relation and the derivatives obtained from it satisfy the given differential equation.

We are now going to discuss the problem of finding general solution (integral) of differential equations. The solution of any differential equation may or may not exist. Even if the integral of the given equation exists, it may not be easy to find. Methods of solutions of following special types of differential equations are chiefly taught;

i) Equations solvable by separation of variables. ii) Homogeneous equations. iii) Linear equations of first order. iv) Exact differential equation.

In this lecture we are to study type (i). SEPERABLE DIFFERENTIAL EQUATIONS:- A differential equation of the type,

F(x) G(y) dx + f(x) g(y) dy =0 (A) Is called equation with variables separable or simply a separable equation. Equation (A) can be written as,

G(+)H(+) IJ + K(*)

L(*) dy = 0

This can be easily integrated. Following are couple of solved problems that certainly going to help getting the idea. Q) y (1+x2)1/2 dy + x ( 1+ y2)1/2 dx = 0 Sol:- ⇒ y √1 + J3 dy = -x P1 + Q3 dx ⇒ *

PRS *0 dy = T+√RS +0 dx

⇒ U *PRS *0 dy = − U +

√RS +0 dx

Let y = tanW ⇒ dy = sec2W

x = tan∅ ⇒ dx = sec2∅

⇒ U YZ[\ ]^_0\ √RS `ab0\ IW = − U YZ[∅ cde0 ∅

√RS `ab0∅ I∅

⇒ U YZ[\ ]^_0\ √cde0\ IW = − U YZ[∅ cde0 ∅

Pcde0∅ I∅ . f 1+ tan2g = sec2g

⇒ U YZ[\ ]^_0\ cde\ IW = − U YZ[∅ cde0 ∅

cde∅ I∅

⇒ U hijW klmW IW = − U hij∅ klm∅ I∅ ⇒ sec W + C1 = - (sec ∅ + C2 ) ⇒ sec W + sec ∅ = -C1 – C2 . f - C1 – C2 = C ⇒ √1 + hij3W + P1 + hij3∅ = C ⇒

Q) ( 1+ x2) dy – xy dx =0 Sol:- ⇒ (1 + x2) dy = xy dx ⇒ Ro dy = +

RS+0 IJ

⇒ U Ro dy = U +RS +0 IJ

Let 1+ x2 = t ⇒ 2x dx = dt ⇒ x dx = R3 Ih

⇒ ln y + C1 = R3 U R` Ih

⇒ ln y + C1 = R3 ln t + C2

⇒ ln y - R3 ln t = C2 – C1 ⇒

OR ⇒ ln y – ln ( 1+ x2)1/2 = ln C

⇒ ln p *√S +0q = ln r

⇒ *√S +0 = C

P1 + Q3 + √1 + J3 = C

ln y - R3 ln(1 + J3) = ln r

⇒ y = C √+ J3 Answer Q) ( 4x+y)2 tu

tv = 1

Sol:- ( 4x+y)2 )+

)* = 1 ----------------- (i)

Let 4x+y=z . On differentiating b/s w.r.t. ‘y’ we get; ⇒ 4 )+

)* + 1 = )z)*

⇒ 4 )+)* = )z

)* – 1 ⇒ )+)* =R

{ 1)z)* – 1 2

Thus;

i) ⇒ z0{ 1)z

)* – 1 2 = 1 ⇒ z2 1)z)* – 1 2 = 4 ⇒ z2 )z

)* - |3 = 4

⇒ z2 )z)* = |3 + 4 ⇒ z0

z0S R dz = dy ⇒ U z0z0S R dz = U IQ

Let z= 2 tanW ⇒ dz = 2 sec2W IW

⇒ U 11 − {z0S R2 dz = U IQ ⇒ U I| − 4 U R

z0S R I| = U IQ

⇒ z - 4 U 3 cde0 \{ `ab0\S{ IW = U IQ ⇒ z – {.3

{ U cde0 \`ab0\SR dW = Q + C1

⇒ z – 2 U cde0 \cde0 \ dW = y + C1 ⇒ z – 2 U IW = y + C1

⇒ z – 2 W + C2 = y + C1 ⇒ z - 2W - y = C1 – C2 ⇒ 4x + y - 2 tan-1 z3 - y = C ⇒ 4x - 2 tan-1 z3 = C

⇒ 2(2x - tan-1 z3 ) = C ⇒ 2x - tan-1 z3 = }3 . f C’ = }3

⇒ 2x – tan-1 {+S*

3 = C’

Q) tu

tv = u ( ~ ��� uS�)��� vSv ��� v

Sol:- ⇒ (sin y + y cos y) dy = x(2 log x + 1) dx ⇒U(sin y + y cos y)dy = U x(2 log x + 1) dx ⇒ U sin Q IQ + U Q cos Q IQ = 2 U J log J IJ + U J IJ Integration by parts ∵ U � � = � � − U ��� Where; f ’ = derivative of “f” & G = integral of “g” f= y , g = cos y f = log x , g= x

f ‘ = 1 , G = sin y f ‘ = R+, , G = +03

⇒ U sin Q IQ + y sin y - U sin Q IQ = 2 ( +03 log x - R3 U +0

3 dx ) + U x dx ⇒ y sin y + C1 = x2 logx – U J IJ + U J IJ + C2 ⇒ y sin y = x2 log x + C2 – C1

⇒ y sin y = x2 log x + C