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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?

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  • $\begingroup$ Every solution of the DE (and you have shown us two) is a "particular solution". Why should different methods give you the same one? As you showed here: they may not give you the same one. In fact, if you use your Wronskian method, and keep the "+C" terms in the integrals, you will get the general solution of the DE. $\endgroup$
    – GEdgar
    Mar 21, 2015 at 1:03
  • $\begingroup$ I guess I'm confused because the textbook says I should check my work by finding the particular solution via both methods. $\endgroup$
    – Frank Gallagher
    Mar 21, 2015 at 1:05
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    $\begingroup$ @FrankGallagher: 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. $\endgroup$
    – Amzoti
    Mar 21, 2015 at 1:06
  • $\begingroup$ Oh thank you! I didn't realize you could add that into the general solution of the homogeneous equation. $\endgroup$
    – Frank Gallagher
    Mar 21, 2015 at 1:08

1 Answer 1

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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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