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I'm having trouble finding the solution for the following differential equation

$ x'' + x = \frac{-2}{cost} $

So the solution of the homogeneous equation is trivial $x_H = c_1\cos(t) + c_2 \sin(t)$

I'm having trouble with the particular solution

How can I pass my $b(t) =\frac{-2}{cost}$ to something like $b(t) = e^{\alpha t}(p_1(t)\cos(\beta t) + c_2(t) \sin(\beta t))$?

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This link seems to describe this well, I have changed to notation to fit your question. You can use a general form for the particular solution when you know the homogeneous part.

You wanted to solve \begin{equation} x''(t) + x(t) = b(t) \end{equation} with \begin{equation} b(t) = \frac{-2}{\cos(t)} \end{equation} and you know \begin{equation} x_c(t) = \underbrace{c_1 \cos(t)}_{x_1(t)} + \underbrace{c_2 \sin(t)}_{x_2(t)} = x_1(t) + x_2(t) \end{equation} define \begin{equation} W(x_1,x_2):=x_1(t)x_2'(t) - x_1'(t)x_2(t) \end{equation} which is called the Wronskian, in this case \begin{equation} W(c_1 \cos(t),c_2 \sin(t))=c_1c_2 \end{equation} then the general solution is \begin{equation} x_p(t) = x_2(t)\int \frac{x_1(t)b(t)}{W(x_1,x_2)}\;dt - x_1 \int \frac{x_2(t)b(t)}{W(x_1,x_2)}\;dt \end{equation} which in this particular case is \begin{equation} x_p(t) = -c_2 \sin(t)\int \frac{2c_1 \cos(t)}{c_1c_2 \cos(t)}\;dt + c_1 \cos(t) \int \frac{2c_2 \sin(t)}{c_1c_2 \cos(t)}\;dt \end{equation} you can cancel the constants and the first integral is easily resolved \begin{equation} x_p(t) = 2\cos(t) \int \tan(t)\;dt -2(t+c_3)\sin(t) \end{equation} \begin{equation} x_p(t) = -2\cos(t) (\log(\cos(t))+c_4) -2(t+c_3)\sin(t) \end{equation} and $x(t)=x_c(t)+x_p(t)$ so \begin{equation} x(t) = c_1 \cos(t) + c_2 \sin(t) -2(t+c_3)\sin(t)-2\cos(t)( \log(\cos(t))+c_4) \end{equation}

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