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$$y''+y=cos3x$$

$$\lambda^2+\lambda=0\Rightarrow \lambda_{1,2}=\pm i$$

$$y_{h}=c_{1}cosx+c_{2}sinx$$

Because the homogeneous solutions is $ c_{1}cosx $ and not $c_{1}cos3x$ we can the particular solution to be:

$$y_{p}=Acos3x+Bsin3x$$

Substituting the particular solution to the ODE:

$$y'_{p}=-3Asin3x+3Bcos3x$$

$$y''_{p}=-9Acos3x-9Bsin3x$$

$$-9Acos3x-9Bsin3x+Acos3x+Bsin3x=cos3x$$

$$-8Acos3x-8Bsin3x=cos3x$$

Is it correct that there is no sin3x on the RHS of the equation? usually it is because the particular solution was incorrect

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    In the last display equation, the second line should read $y_p'' = \cdots$. – 2017-02-06

1 Answers 1

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You're good! Your last line proves that $B = 0$ and $A = -1/8$, so we are saying that one particular solution is $-\frac{1}{8}\cos(3x)$. That is indeed a particular solution (plug it back into the original equation), so you're correct.

It probably made you uncomfortable that $B$ was zero, but that's OK; it corresponds to the fact that the left hand side of your differential equation didn't have any multiple of $y'$ in it, and the right hand side didn't have any sine functions.