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I am trying to look for this question on the internet and Mathstack but I cannot find anything for second order inhomogeneous DE. And I need to know the solution using integrating factor method.

$$y''+2y'-8y=e^{2x}$$

Any links or solution drafts would be appreciated.

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2 Answers 2

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I assume that the idea is to use the integrating factor method on the two first order ODEs that you get out of operator factorization, also known as the annihilator method. Thus you write $D^2+2D-8=(D+4)(D-2)$ where $D$ is the derivative operator. Then you either let $u=y'+4y$ or $u=y'-2y$. In the former case you have $u'-2u=e^{2x}$, in the latter case you have $u'+4u=e^{2x}$. Then you go back and solve for $y$ knowing $u$.

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  • $\begingroup$ Honestly I don't understand how did you get $u'-2u=e^{2x}$. And how should I get integrating factor from those equations? I also not pretty sure if this is the method I need. $\endgroup$
    – Karagum
    Oct 8, 2019 at 10:01
  • $\begingroup$ @Karagum If $u=y'+4y$ then $u'-2u=y''+4y'-2y'-8y=y''+2y'-8y$. I am not totally clear on what you meant, though. The next thing that comes to mind as something integrating-factor-esque is variation of parameters, but no one calls that qn integrating factor methos to my knowledge even though it kind of is one, at least in regards to its derivation. $\endgroup$
    – Ian
    Oct 8, 2019 at 10:10
  • $\begingroup$ Alright, it looks better now. And what if I have a repeated root? For example $y''+6y'+9y=e^{-3x}$? Then I get D=-3 for both cases. $\endgroup$
    – Karagum
    Oct 8, 2019 at 10:19
  • $\begingroup$ @Karagum The method works the same, $u=y'+3y$ and $u'+3u=e^{-3x}$. You just don't have as much freedom to pick here. In this method the extra factors of $x$ show up as a result of how the integrals work out. $\endgroup$
    – Ian
    Oct 8, 2019 at 10:21
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This second order inhomogeneous ODE can be written using $\frac{d}{dx}=D$ as $$(D^2+2D-8)y=e^{2x} \implies (D+4)(D-2)y=e^{2x}.$$ Let $(D-2)y=u$, then we get $$Du+4u=e^{2x}\implies \frac{du}{dx}+4u=e^{2x}.$$ The integrating factor for this first order ODE is $e^{4x}$, so $$u(x)=e^{-4x} \int e^{2x} e^{4x}+C_1 e^{-4x} \implies u(x)=\frac{1}{6} e^{2x}+C_1 e^{-4x}.$$ Next, we consider $(D-2)y=u$, which is $$\frac{dt}{dx}-2y=\frac{1}{6}e^{2x}+C_1 e^{4x}$$ For this ODE the integrating factor is $e^{-2x}$, then we get $$y=e^{2x} \int (\frac{1}{6} e^{2x}+C_1 e^{-4x}) e^{-2x} dx+C_2 e^{2x} \implies y=\frac{1}{6}xe^{2x}-\frac{C_1}{6} e^{-4x}+C_2 e^{2x}$$. Finally, the final solution of the given second order ODE is $$y=\frac{1}{6}xe^{2x}+C_3 e^{-4x}+C_1 e^{2x,}$$ where $C_1,C_3$ are constants independent of $x$, the first part is called particular solution and the rest is called tge general solution.

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