- #1

Kaguro

- 221

- 57

## Homework Statement

Solve the following DE with the method of undetermined coefficients:

y'' + 4y = 2cos(3x)cos(x)

## Homework Equations

2cos(3x)cos(x) = cos(4x) + cos(2x)

## The Attempt at a Solution

Let's split the particular integral into two parts: yp1 and yp2.

So yp1 is solution for RHS=cos(4x) and yp2 is solution for RHS=cos(2x)

Let yp1 = Acos4x + Bsin4x

(yp1)'' = -16Acos(4x) -16Bsin(4x)

So, -16Acos(4x) - 16Bsin(4x) + Acos(4x) + Bsin(4x) = cos(4x)

So by equating coefficients, we have:

yp1 = (-1/12)cos(4x)

In case of yp2, I have taken:

Let yp2 = Acos(2x) + Bsin(2x)

yp2'' = -4Acos(2x) -4Bsin(2x)

So 0 = cos(2x)

That's not great...