Lecture 11 - Jan 29 (extra)

29/1/26

Proof

Let be solutions of . Assume that is a solution of for some functions of .

We further assume that .

Then,

Since is a solution of (2),

Therefore,

As are solutions, the first two terms are zero.

Thus, we have and .

Solving for and ,

Where .

---

For solving

• Step 1: Find two solutions of .
• Step 2: Find a particular solution of .
• Step 3: The general solution .
  • and .

Example.

Find a particular solution of:

Exercise.

Solve , .

Solution 1.

.

Consider .

By inspection, is a solution.

Solution 1 (Continued).

To find the second solution, we use the transformation :

Thus, we have and , and and .

The Wronskian is:

Step 2: To find the particular solution

---

Solution 2.

.

For the homogeneous equation , the characteristic equation is:

The homogeneous solutions are:

The Wronskian is:

For the particular solution :

Theorem

Let , where and are continuous functions. If are a set of linearly independent (LI) solutions of the corresponding homogeneous equation, then a particular solution of is given by

where is defined by

where is the Wronskian of and is the determinant obtained from by replacing the column

Proof

Let be the solutions of the homogeneous equation. Put and suppose .

Consider .

We assume that .

Then, .

.

Again assume .

Proceeding in this way, we have

Accordingly, we have

Then .

From (A), we need to solve for . By Cramer’s rule: