Use reduction of order to find a solution of the given nonhomogeneous equation. The indicated function $y_1(x)$ is a solution of the associated homogeneous equation. Determine a second solution of the homogeneous equation and a particular solution of the nonhomogeneous equation.

$$ y''-4y=2; y_1=e^{-2x} $$

I am very confused on their free use of "homogeneous" and "nonhomogeneous". This is how I am understanding the question:

"the given nonhomogeneous equation": $y''-4y=2$

"the associated homogeneous equation": ???

My teacher told me that a homogeneous equation in this context is one that equals zero, however there is none here. also, when I plug $y_1$ into $y''-4y=2$ it is not valid because you get $0=2$.

  • $\begingroup$ $\color{blue}{\text{The associated homogeneous equation: }y''-4y=0}$. Note that if $\color{blue}{\varphi(x)}$ is a solution to the last homogeneous equation, then $\color{blue}{y=\varphi(x)-\frac{1}{2}}$ is a solution for $$y''-4y=2$$ $\endgroup$ – Ángel Mario Gallegos Jun 28 '15 at 19:09

The homogeneous equation associated with $y''-4y=2$ is $y''-4y=0$. Once you know the solutions $y_1$, $y_2$ of the homogeneous equation, you can easily find the particular solution using the method of variation of parameters: $y(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$. See here for more.

  • $\begingroup$ Or just observe that $y=1/2$ is a particular solution, $\endgroup$ – Matematleta Jun 28 '15 at 19:22
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    $\begingroup$ Thats a lot quicker. Of course you mean $y=-1/2$ $\endgroup$ – zed111 Jun 28 '15 at 19:25
  • $\begingroup$ yes, I did thanks! $\endgroup$ – Matematleta Jun 28 '15 at 19:27
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    $\begingroup$ We haven't learned variation of parameters so I don't know what my teacher will think of if he sees me using this. Is there another way? $\endgroup$ – Aaron Jun 28 '15 at 19:45
  • $\begingroup$ In that case, use Chilango's comment $\endgroup$ – zed111 Jun 28 '15 at 19:49

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