Jan 29, Lect - 10

Problem:

Solve:

  1. , if is one of the solutions.
  2. .

Solution 1.

.

The solution,

Solution 2.

By inspection, () is a solution.


Theorem

Let be a non-trivial solution of

, for any , and are continuous functions on .

Then the transformation reduces the equation into an order equation.

when .

If are linearly independent (LI) solutions of (2) and if for , then

is a basis of solutions of (1), where .

Proof

Let .

Suppose is a solution of (1), then

Sorting according to :

when .

This is an order Differential Equation (DE) and it has LI solutions, .

Since , .

Therefore are the solutions of (1).

It remains to show that the above solutions form a basis.

Suppose .

since are LI.

, and hence for all .

are LI and form a basis.

And if for , then

is a basis of solution of (1), where

Method of Variation of Parameters

Theorem

Let

where , , and are continuous functions of defined on an interval . If are two linearly independent solutions corresponding to the homogeneous equation,

then, a particular solution of (1) is given by,

where

where is the Wronskian of .

This is an order differential equation and it has linearly independent solutions namely .

Since .

Therefore, are solutions of (1). It remains to show that the above solution is a basis.

  • Suppose
    • since
    • (differentiating)
    • for since are linearly independent.
    • and hence for all .
    • are linearly independent and hence a basis.

Method of Variation of Parameters

Theorem

Set where , , and are continuous functions of defined on an interval . If are two linearly independent solutions corresponding to the homogeneous equation , then a particular solution of (1) is given by where

where is the Wronskian of .