MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai 76dey@math.iitb.ac.in

March 10, 2014

Santanu Dey Lecture 7

Outline of the lecture

Lipschitz continuity

Existence & uniqueness

Picard’s iteration

Second order linear equations

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.

1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) =

1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt =

1 +x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) =

1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt

= 1 +x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) =

1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n.

(By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) =

limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x)

= ex2/2.

Santanu Dey Lecture 7

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 7

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?

(No, consider f (x) =√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 7

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 7

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .?

(n=3).

Santanu Dey Lecture 7

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 7

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt

(1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt.

Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0,

U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (1)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (1) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 7

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 7

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0

=⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 7

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0

=⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 7

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 7

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations

- Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separable

Reducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separable

Exact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factors

Reducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theorem

Picard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 7

Second order differential equations

Recall that a general second order linear ODE is of the form

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.

Santanu Dey Lecture 7

Second order differential equations

Recall that a general second order linear ODE is of the form

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.

Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.

Santanu Dey Lecture 7

Second order differential equations

Recall that a general second order linear ODE is of the form

a2(x)d2y

dx2+ a1(x)

dy

dx+ a0(x)y = g(x).

An ODE of the form

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x)

is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.

Santanu Dey Lecture 7

Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.

Santanu Dey Lecture 7

Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.

Santanu Dey Lecture 7

Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.

Otherwise it is called non-homogeneous.

Santanu Dey Lecture 7

Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.

Santanu Dey Lecture 7

Homogeneous Linear Second Order DE

If r(x) ≡ 0 in the equation

d2y

dx2+ p(x)

dy

dx+ q(x)y = r(x),

that is,d2y

dx2+ p(x)

dy

dx+ q(x)y = 0,

then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.

Santanu Dey Lecture 7

Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .

Santanu Dey Lecture 7

Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0,

y(x0) = a, y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .

Santanu Dey Lecture 7

Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a,

y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .

Santanu Dey Lecture 7

Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .

Santanu Dey Lecture 7

Initial Value Problem- Existence/Uniqueness

An initial value problem of a second order homogeneous linearODE is of the form:

y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,

where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .

Santanu Dey Lecture 7