Lecture 08 - Jan 22

Jan 22, Lect - 8

Uniqueness Theorem for IVP

Let be any two constants and on any interval containing . Then there exists at most one solution of the Initial Value Problem (IVP):

Proof

Let and be the two solutions of:

is a solution of (1). .

is a solution of (1). .

Let .

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Problems

Example 1.

Solve

Example 2.

Consider the equation

a) Compute the solution satisfying . b) Compute the solution satisfying . c) Compute and .

Solution 1.

Characteristic equation:

Solution:

Solution 2.

Characteristic equation:

General solution:

For :

Similarly for :

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order homogeneous linear equation with constant coefficients

Theorem

Consider the order linear equation with constant coefficients,

i) If are distinct roots of the characteristic polynomial , then

are linearly independent (LI) solutions of .

ii) If be distinct roots of the characteristic polynomial , and each is of multiplicity , with .

Then,

are LI solutions.

Proof

Consider .

When .

(i) If has distinct roots , then are the solutions of .

(ii) Suppose is a root of with multiplicity . .

Claim

is a solution of for .

Consider,

But,

For :

This is true for any .

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Theorem

Theorem

Let be the solution of

on an interval containing a point . Then for all ,

where .

Proof

Let .

Since is a solution of , we have

Substitute in (A):

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