# Setting up a particular equation with (cos(x))^2

## Homework Statement

Find a particular solution yp of the differential equation

2y''+9y'+2y=(cos(x))^2

## The Attempt at a Solution

I'm not really sure where to start? I solve for the homogeneous solution and get C1e^ -4.26556+C2e^-.2344355629=0. I'm just not sure where to go from there. Should I do
y(x)=(Acos(x)+Bsin(x))^2 since it's just cos (x)*cos(x) or is not that right approach?

$$cos^{2}(x)$$ can be broken up using the half angle formula. You'd get a polynomial (degree one) and a trig function that is expandable under Euler's Formula. Add the particular solutions together to get the final particular solution.

so (cos(x)^2) is ((1+cos (2x))/2.
Which is 1/2 + cos(2x)/2. Am I missing something to get it into a degree one term and a trig function?
For that I get A+ B*cos(2x)+C*sin(2x)?

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so (cos(x)^2) is ((1+cos (2x))/2.
Which is 1/2 + cos(2x)/2. Am I missing something to get it into a degree one term and a trig function?
For that I get A+ B*cos(2x)+C*sin(2x)?

Sorry, not degree one. I meant it has a constant added to a trig function. But you can solve the equation and get particular solutions by guessing that $$y = \frac{1}{4}$$ (for one solution) and that $$y = Re(Ae^{2i\theta})$$ where A is some constant. Or you can use your method and simply guess $$A sin(2\theta)+Bcos(2\theta)$$. Add that guess to your other particular solution ($$y = \frac{1}{4}$$), then add your particular solutions to your homogeneous solution. This will result in the general solution of the differential equation.

Ah ok, I don't even need the general solution, just the particular solution, so should I just consider the two particular solutions of Asin(2x)+Bcos(2x) and y=1/4 (Solving the constants A and B?)

Ah ok, I don't even need the general solution, just the particular solution, so should I just consider the two particular solutions of Asin(2x)+Bcos(2x) and y=1/4 (Solving the constants A and B?)

Yup. Add the two particular's together and you're done!