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.
On
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 $u$.
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.