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I tried using the method of undetermined coefficients to find the particular solution for this equation. However, I got stuck halfway through.

Solution

I'm not sure if this is the correct way to approach this equation. Help will be greatly appreciated.

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    $\begingroup$ That doesn't look like undetermined coefficients - at least not the way I learned it. Why not set y = Acos(3x) + Bsin(3x), differentiate twice and plug into the original equation? $\endgroup$ Oct 30 '20 at 14:53
  • $\begingroup$ Welcome to the website. Please use Mathjax to format your equations in future, instead of attaching pictures. $\endgroup$ Oct 30 '20 at 15:06
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You have solved it correctly. You get $U=145/(12i-1)=-1-12i$, so the paeticular solution of $y''+4y'+8y=145e^{i3x}$ is $Ue^{i3x}=(-1-12i)e^{i3x}=12\sin3x-\cos3x+i(\cdot\cdot\cdot)$. Clearly the real part of this particular solution generates the real part of the ODE it satisfies i.e. $145\cos3x$, so the particular solution of original ODE is $12\sin3x-\cos 3x$.


Suppose $f(x)+ig(x)$ is a solution of $\mathcal Ly=h(x)+ik(x)$, where $\mathcal L$ is a linear differential operator with real-valued coefficients (in your case $D^2+4D+8)$. Then$$\mathcal L(f+ig)=\cal Lf+i~\cal Lg=h+ik$$Now equate the real and complex parts. This means $f$ satisfies $\cal Lf=h$ and $g$ satisfies $\cal Lg=k$.

In your case, $f(x)=12\sin3x-\cos 3x$ and $h(x)=145\cos3x$.

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