MA-108 Ordinary Di erential Equationskeshari/D3-Lecture-3.pdf · 2015. 3. 9. · 13 14D260002 KANNAN KARTHIK RAJAN 14 140100036 NISHANT NEERAJ 15 140260018 RAKSHIT JAIN 16 14B030029

MA-108 Ordinary Differential Equations

M.K. Keshari

Department of MathematicsIndian Institute of Technology Bombay

Powai, Mumbai - 76

9th March, 2015D3 - Lecture 3

M.K. Keshari D3 - Lecture 3

Solution of 1st order linear ODE

Theorem (Existence and Uniqueness for 1st order linear ODE)

Assume p(x) and f(x) are continuous on I = (a, b).If x0 ∈ I, then IVP

y′ + p(x)y = f(x), y(x0) = y0 ∈ R

has a unique solution y = φ(x) on the interval I.

The solution of the homogeneous part y′ + p(x)y = 0 is givenby

y1(x) = e−∫p(x) dx

By variation of parameters method, the solution of ODE isy = uy1, where u′(x) = f(x)/y1(x). Therefore

y(x) = e−∫p(x) dx

(∫f(x) e

∫p(x) dx dx+ C

)M.K. Keshari D3 - Lecture 3

Solution of 1st order linear ODE

• The interval of existence and uniqueness of the solution ofIVP

y′ + p(x)y = f(x), y(x0) = y0 ∈ R

is independent of y0.

• The uniqueness condition implies that the one parameterfamily of solutions,

y(x) = e−∫p(x) dx

(∫f(x) e

∫p(x) dx dx+ C

)is, in fact, a general solution on the interval I.

• Any solution of the IVP is obtained from the generalsolution for some scalar C.


Interval of validity of an ODE

A set S ⊂ R is open if for any x ∈ S, there exist ε > 0 suchthat (−ε+ x, x+ ε) ⊂ S.

An open set S ⊂ R is connected if α, β ∈ S and x ∈ R suchthat α < x < β, then x ∈ S.

A connected open sets in R is called an (open) interval andthey are of the form (a, b) for −∞ ≤ a < b ≤ ∞.

Let Ω = (0, 1) ∪ (1, 2). Then Ω is not connected. It is unionof two disjoint open intervals. Let

f : (0, 1)→ R and g : (1, 2)→ R

be n-times differential functions. Define

h : Ω→ R as h|(0,1) = f and h|(1,2) = g

Then h is n-times differential function.M.K. Keshari D3 - Lecture 3

Interval of validity is a connected open interval

Assume the general solution of

y′ = f(x, y)

is defined on the interval Ω = (0, 1) ∪ (1, 2). Let x0 ∈ (0, 1)and x1, x2 ∈ (1, 2). Let

y0(x) : (0, 1)→ R

be a solution of y′ = f(x, y) with initial condition y(x0) = a0.Let

y1(x), y2(x) : (1, 2)→ R

be solutions of y′ = f(x, y) with initial conditions

y(x1) = a1 and y(x2) = a2

respectively.M.K. Keshari D3 - Lecture 3

Interval of validity is a connected open interval

Let us defineh1(x), h2(x) : Ω→ R as

h1(x) = h2(x) = y0(x) for x ∈ (0, 1) and

h1(x) = y1(x), h2(x) = y2(x) for x ∈ (1, 2)

Then h1 and h2 are solutions of y′ = f(x, y) onΩ = (0, 1) ∪ (1, 2) with same initial condition y(0) = a0.

Remark. Therefore, uniqueness of solution of an IVP:

y′ = f(x, y), y(x0) = a0

means there exist a unique solution on a (connected) openinterval containing x0.The interval of validity for a solution of an IVP will always bean open interval (connected).


Example

Solvey′ + (cotx)y = x cscx

The functions p(x) = cot x and f(x) = x cscx are bothcontinuous except at points x = nπ for integers n.Let’s find solutions of ODE on the intervals (nπ, (n+ 1)π).A solution of homogeneous part is

y1(x) = e−∫p(x) dx = e−

∫cotx dx = e− ln | sinx| =

1

sinx

Therefore the solution of ODE is y(x) =

y1(x)

(∫f(x)

y1(x)dx+ C

)=

1

sinx

(∫x cscx sinx dx+ C

)

=1

sinx

(∫x dx+ C

)=

1

sinx

(x2

2+ C

)M.K. Keshari D3 - Lecture 3

Example continued..

If we put the initial condition y(π/2) = 1 in the generalsolution

y(x) =1

sinx

(x2

2+ C

)then

1 =π2

8+ C =⇒ C = 1− π2

8

Thus the solution of IVP is

y(x) =x2

2 sinx+

(1− π2

8)

sinx

The interval of validity of this solution is (0, π).


Solution in terms of integral

Example. Solve the IVP y′ − 2xy = 1, y(0) = y0The solution of the homogeneous part is

y1(x) = e∫−p(x) dx = e

∫2x dx = ex

2

The general solution is

y(x) = y1(x)

(∫f(x)

y1(x)dx+ C

)= ex

2

(∫e−x

2

dx+ C

)Since the initial is at x0 = 0, rewrite the general solution as

y(x) = ex2

(∫ x

0

e−t2

dt+ C

)y(0) = y0 gives C = y0. The IVP has solution (defined on R)

y(x) = ex2

(∫ x

0

e−t2

dt+ y0

)


1 140100003 YADAV RAM RAMASHANKAR2 140100011 TANMAY3 140100021 PRANIT BHANDARI4 140100043 AMIT A PATIL5 140100057 HARVEER SINGH6 140100077 HIMANSHU DENGREABSENT7 140100090 NAVNEET AGARWAL8 140100105 MANDA PUNEETH9 140260004 VARUN SREEDHARA BHATT10 14B030001 KARAN MEHTA11 14B030011 VYAPARI APURV ATUL12 14B030028 ASTIK ANAND13 14D260002 KANNAN KARTHIK RAJAN14 140100036 NISHANT NEERAJ15 140260018 RAKSHIT JAIN16 14B030029 JOHN LAZARUS17 140100004 CHAUDHARI AKUL ANANT18 140100014 SHANBHAG SOHAM SUBHASH19 140100026 KANSARA VATSAL SANJAY


Solution of 1st order Non-Linear ODE

Existence and Uniqueness for Non-Linear ODE.

(a) (Existence) Assume f(x, y) is continuous on an openrectangle R := (x, y) ∈ R2| a < x < b, c < y < d that

contains (x0, y0). Then IVP: y′ = f(x, y), y(x0) = y0 (∗)has at least one solution on some interval (a1, b1) ⊂ (a, b)containing x0.

(b) (Uniqueness) If both f and ∂f/∂y are continuous on R,then IVP (*) has a unique solution on some interval(a′, b′) ⊂ (a, b) containing x0.

Remark. (a) is an existence theorem. It guarantees asolution on some interval containing x0, but does not give anyinformation on how to find the solution or how to find theinterval of validity. In this case, IVP can have more than onesolution.



(b) is a uniqueness theorem. It guarantees that IVP has aunique solution on some interval (a′, b′) ⊂ (a, b) containing x0.However, if (a′, b′) 6= (−∞,∞), then IVP may have more thanone solution on a larger interval containg (a′, b′).

For example. it may happen that b′ <∞ and two solutionsy1, y2 are defined on some interval (a′, b1) with b1 > b′, andhave different values for b′ < x < b1.

Thus the graph of y1 and y2 branch off in different directionsat x = b′.

In this case, since y1 = y2 on (a′, b′), by continuity,y1(b

′) = y2(b′) := y.



Now y1 and y2 are both solutions of the IVP:

y′ = f(x, y), y(b′) = y (∗∗)

they differ on every open interval containing b′.

Therefore, f or ∂f/∂y must have a discontinuity at somepoint in each open rectangle that contains (b′, y) ∈ R2.

Why?

If not, then by uniqueness theorem, (**) will have a uniquesolution on some open interval containing b′, a contradiction.


Example

Ex. Consider the IVP

y′ =x2 − y2

1 + x2 + y2, y(x0) = y0 (∗)

If

f(x, y) =x2 − y2

1 + x2 + y2, then

∂f

∂y=

−2y

1 + x2 + y2+−2y(x2 − y2)(1 + x2 + y2)2

=−2y(1 + 2x2)

(1 + x2 + y2)2

Since f(x, y) and ∂f/∂y are continuous for all (x, y) ∈ R2,by existence and uniqueness theorem, if (x0, y0) is arbitrary,then (∗) has a unique solution on some open intervalcontaining x0.


Ex. Consider the IVP y′ =x2 − y2

x2 + y2, y(x0) = y0 (∗)

If f(x, y) =x2 − y2

x2 + y2, then

∂f

∂y=−2y

x2 + y2+−2y(x2 − y2)

(x2 + y2)2=−4x2y

(x2 + y2)2

Here f(x, y) and ∂f/∂y are continuous for all (x, y) ∈ R2,except at (0, 0).

If (x0, y0) 6= (0, 0), then there is an open rectangle Rcontaining (x0, y0) but not containing (0, 0).

Since f(x, y) and ∂f/∂y are continuous on R, by existenceand uniqueness theorem, if (x0, y0) 6= (0, 0), then (∗) has aunique solution on some open interval containing x0.


Example


y′ =x+ y

x− y, y(x0) = y0 (∗)

If

f(x, y) =x+ y

x− y, then

∂f

∂y=

2x

(x− y)2

Here f(x, y) and ∂f/∂y are continuous everywhere except onthe line y = x.

If x0 6= y0, there is an open rectangle R containing (x0, y0)that does not intersect with the line y = x.Since f(x, y) and ∂f/∂y are continuous on R, by existenceand uniqueness theorem, if x0 6= y0, then (∗) has a uniquesolution on some open interval containing x0.



y′ =10

3xy2/5, y(x0) = y0 (∗)

Q1. For what (x0, y0), (∗) has a solution?Q2. For what (x0, y0), (∗) has a unique solution on someopen interval that contains x0?

Here f(x, y) =10

3xy2/5 and

∂f

∂y=

4

3xy−3/5

• Since f(x, y) is continuous for all (x, y) ∈ R2, IVP (∗) has asolution for all (x0, y0).

• Since f(x, y) and ∂f/∂y both are continuous for all (x, y)with y 6= 0. If y0 6= 0, there is an open rectangle R containing(x0, y0) s.t. f and ∂f/∂y are continuous on R. Hence IVP (∗)has a unique solution on some open interval containing x0.


Ex. Consider f(x, y) =10

3xy2/5 and ∂f/∂y =

4

3xy−3/5.

Since ∂f/∂y is not defined for y = 0, it is discontinuous ify = 0. Hence the uniqueness theorem does not apply wheny0 = 0.Hence the IVP

y′ =10

3xy2/5, y(0) = 0 (∗)

may have more than one solution on every open intervalcontaining x0 = 0. We will show that this is true!

By inspection y ≡ 0 is a solution of (∗).

We will show that there are non-zero solutions to the IVP (*).

Let y be a non-zero solution of y′ =10

3xy2/5.


Example continued ...

Then separating variables, we get

y′

y2/5=

10

3x

Integrating it, we get

5

3y3/5 =

5

3(x2 + C)

=⇒ y(x) = (x2 + C)5/3 (∗∗)Since we divided by y2/5 to separate variables this solution islegitimate only on the open intervals where y(x) does not takezero values.

However, (**) is defined for all (x, y). Differentiating it, we get

y′ =5

3(x2 + C)2/3(2x) =

10

3xy2/5, ∀x ∈ (−∞,∞)


Example continued ...

Thus

y(x) = (x2 + C)5/3 satisfies y′ = (10/3)y2/5

on (−∞,∞) for all C.

Now y(0) = 0 in y(x) = (x2 + C)5/3 gives C = 0.

Thus y(x) = x10/3 is another solution of IVP (*).

Thus we have two solutions of IVP

y′ = (10/3)y2/5, y(0) = 0 (∗)

namely y ≡ 0 and y(x) = x10/3.

We can construct two more solutions of IVP (*). How?


Ex. The IVP y′ =10

3xy2/5, y(0) = −1 (∗)

has a unique solution on some open interval containingx0 = 0.

Find a solution and the largest open interval (a, b) on whichthis solution is unique.

Let y(x) be any solution of IVP. Since y(0) = −1, there is anopen interval I containing x0 = 0 such that y takes non-zerovalues on I. In this case, the general solution of ODEy′ = (10/3)xy2/5, is given by (solved earlier)

y(x) = (x2 + C)5/3

From y(0) = −1, we get C = −1. Hence

y(x) = (x2 − 1)5/3 for x ∈ I (∗∗)

Note y(x) = (x2 − 1)5/3 is a solution of y′ = (10/3)xy2/5,which is defined on (−∞,∞).


Hence every solution of (*), which does not take zero-value, isgiven by y(x) = (x2 − 1)5/3 on open interval (−1, 1).

This is the unique solution of IVP (*) on (−1, 1). Why? UseExistence and Uniqueness theorem for all x0 ∈ (−1, 1) andinitial condition y(x0) = (x20 − 1)5/3.

(−1, 1) is the largest interval on which IVP (*) has a uniquesolution. To see this, note that we can define another solution

y1(x) =

(x2 − 1)5/3 , −1 ≤ x ≤ 1

0 , |x| > 1

This also shows that the largest interval on which the solutionof (*) is unique is (−1, 1). This solution can be extended onlarger interval (−∞,∞) by y1 and y both.

Exercise. Find largest interval where y′ =10

3xy2/5, y(0) = 1

has a unique solution.


Transforming Non-Linear into Separable ODE

A non-linear differential equation y′ + p(x)y = f(x)yr (∗),where r ∈ R− 0, 1 is said to be a Bernoulli Equation. Forr = 0, 1, it is linear.

If y1 is a non-zero solution of y′ + p(x)y = 0, then puttingy = uy1 in (*), we get

u′y1 + uy′1 + puy1 = furyr1

=⇒ u′y1 = furyr1

=⇒ u′

ur= f(x)(y1(x))r−1

=⇒ u−r+1

−r + 1=∫f(x)(y1(x))r−1 dx+ C.

Ex. Solve y′ − y = xy2.


Converting Non-Linear into Separable ODE

Consider y′ = f(x, y).Substitute y = uy1, where y1(x) is known function and u(x)unknown.

u′y1(x) + uy′1(x) = f(x, uy1(x)),

=⇒ u′y1(x) = f(x, uy1(x))− uy′1(x).

If f(x, uy1(x)) = q(u)y′1(x) for some function u, then

u′y1(x) = (q(u)− u)y′1(x) is separable.

After checking for constant solutions u = u0 s.t. q(u0) = u0,we can separate variables to obtain

u′

q(u)− u=y′1(x)

y1(x)


Homogeneous Non-Linear Equations

Def. A differential equation y′ = f(x, y) is said to behomogeneous if it can be written as y′ = q(y/x).

Substitute y = vx, where v is an unknown function, we get

v′x+ v = q(v) a separable ODE.

Example. Solve xy′ = y + x (∗).

Rewrite it as y′ =y

x+ 1. This is homogeneous ODE.

Substitute y = vx. We get v′x+ v = v + 1 =⇒ v′x = 1.

By integration, v(x) = ln |x|+ C.

Thus the solution to (*) is y = x(ln |x|+ C).


Example: Solve

x2y′ = y2 + xy − x2

Write the ODE as

y′ =y2 + xy − x2

x2=(yx

)2+y

x− 1

This is homogeneous. Substitute y = vx to get

v′x+v = v2+v−1 =⇒ v′

v2 − 1=

1

2

(1

v − 1− 1

v + 1

)v′ =

1

x

Integration gives

1

2(ln |v − 1| − ln |v + 1|) = ln |x|+ C1

=⇒ v − 1

v + 1= Cx2 =⇒ v =

1 + Cx2

1− Cx2


ODE y′ = (y/x)2 + (y/x)− 1

Therefore

y = x1 + Cx2

1− Cx2Q. Are these all the solutions? No!

Both y = x and y = −x are also solutions, but only y = x canbe obtained from the general solution.

The solutions y = x1 + Cx2

1− Cx2were obtained in the intervals

not containing 0.

Does this mean that the only solutions to the ODE, in aninterval containing zero, are y = x or y = −x?


Interval of Validity

Note that y = x1 + Cx2

1− Cx2is continuous at x = 0 and satisfies

the ODE x2y′ = y2 + xy − x2 trivially at that point, sincey(0) = 0.In fact, for arbitrary C1, C2 ∈ R, the function

y(x) =

x

1 + C1x2

1− C1x2if x < 0

x1 + C2x

2

1− C2x2if x ≥ 0

is differentiable and satisfies the ODE x2y′ = y2 + xy − x2with y(0) = 0. Thus this IVP has infinitely many solutions onefor each choice of C1, C2.

We have noted before that the interval of validity will dependon C and hence on the initial condition.


Documents

MA-108 Ordinary Di erential Equationskeshari/D3-Lecture-3.pdf · 2015. 3. 9. · 13 14D260002 KANNAN KARTHIK RAJAN 14 140100036 NISHANT NEERAJ 15 140260018 RAKSHIT JAIN 16 14B030029