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Find particular solution of $\mathrm{y}'' + 3\mathrm{y}' + 2\mathrm{y} = 4\,\mathrm{e}^{-3t}$ using variation of parameters.

I already found one using undetermined coefficients, and got $\,\mathrm{y_{p}} = 2\,\mathrm{e}^{-3t}$. However, I'm getting a particular solution of $\,\mathrm{y_{p}} = \,\mathrm{e}^{-3t}$. How do I show that the general forms of these two are in fact equal ?.

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For the homogeneous differential equation $$y''+3y'+2y=0$$ the characteristic polynomial $r^2+3r+2=0$ has solutions $r=-1$ and $r=-2$ which means that$$y''+3y'+2y=0\implies y=c_1e^{-t}+c_2e^{-2t}$$ Now, for the complete equation, you could have defined $y=e^{-3t}u$ which would lead to $$u''-3u'+2u=4$$ which clearly shows a particular solution $u=2$ which implies $y_p=2e^{-3t}$.

All of that makes $$y''+3y'+2y=4e^{-3t}\implies y=c_1e^{-t}+c_2e^{-2t}+2e^{-3t}$$

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