$\begingroup$The particular solution to $y''-y'-2y=2e^{-t}$ found by the method of undetermined coefficients is $Y(t)=-\frac{2}{3}te^{-t}$
However, when I find the particular solution via the method of variation of parameters I obtain a different expression. I would really appreciate someone to point out where my calculations go awry.
First I find the fundamental set of solutions for the analogous homogeneous equation $y''-y'-2y=0$ which are $y_1(t)=e^{-t}$ and $y_2(t)=e^{2t}$. Then I use these to find the Wronskian $W(y_1,y_2)(t)=3e^t$. Then I plug these values into the following equation to get the particular solution:
$$Y(t)=-y_1(t)\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}dt+y_2(t)\int\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}dt$$ Where $g(t)=2e^{-t}$ is the non-homogeneous term.
$$Y(t)=-e^{-t}\int\frac{e^{2t}2e^{-t}}{3e^t}dt+e^{2t}\int\frac{e^{-t}2e^{-t}}{3e^t}dt$$ $$=-e^{-t}\int\frac{2e^t}{3e^t}dt+e^{2t}\int\frac{2e^{-2t}}{3e^t}dt$$ $$=-e^{-t}\int\frac23dt+e^{2t}\int\frac23e^{-3t}dt$$ Finally giving: $$Y(t)=-\frac23te^{-t}-\frac29e^{-t}$$
So why doesn't this particular solution match the one obtained from undetermined coefficients?
$\endgroup$41 Answer
$\begingroup$Notice that you have as part of the homogeneous solution $c_1 e^{-t}$.
The $-\dfrac29e^{-t}$ gets added to that and just called $c_1 e^{-t}$.
After all, the $-\dfrac 29$ is also just a constant.
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