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Here's the equation I'm working with: y''+y'-2y = 3e^(x)+4x

I want to use the method of undetermined coefficients for this equation. The logical choice for our guess would be y = Ae^(x)+Bx+C.

After substituting in, we end up getting Ae^(x) + Ae^(x)+B - 2(Ae^(x)+Bx+C) = 3e^(x)+4x.

While we get a value for B, -2, it doesn't quite work out for A and C. Thoughts?

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  • $\begingroup$ y = c1e^(-2x)+c2e^(x) $\endgroup$ – zthomas.nc Oct 20 '14 at 0:43
  • $\begingroup$ See this answer. $\endgroup$ – Git Gud Oct 20 '14 at 1:04
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Observe that, if $y=Ae^x$, then $y''+y'-2y=Ae^x+Ae^x-2Ae^x=0$. Therefore, the $Ae^x$ term in your so-called "logical guess" is not contributing anything. That's why you should solve the homogeneous equation first, before you try to find a particular solution. The general solution of the homogeneous equation is $y=c_1e^x+c_2e^{-2x}$, and you can see from that $c_1e^x$ term that $Ae^x$ is useless in guessing a particular solution. Hint: Try replacing $Ae^x$ with $Axe^x$.

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When you said A doesn't work it, it's because A disappears from the equation, correct? A disappears because $e^x$ is a solution of the homogeneous equation. I assume you know how to resolve this problem.

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  • $\begingroup$ ... That is true, it is a homogeneous solution. If a solution appears twice .... multiply by x? And then add xe^(x) as a solution? $\endgroup$ – zthomas.nc Oct 20 '14 at 0:42

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