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I'm struggling solving this linear ODE:

$$y''(x) + 2y'(x) = -4$$

I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;

But I'm struggling with the non-homogeneous part: If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.

What am I doing wrong! Thanks a lot!

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  • $\begingroup$ You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$. $\endgroup$ – Christoph Jan 20 at 9:38
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Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form $$y_p(x)=x\cdot C$$ where $C$ is a constant to be determined.

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  • $\begingroup$ Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :) $\endgroup$ – scalpula Jan 20 at 12:11

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