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So I have the following question here.

Suppose that $y_1$ solves $2y''+y'+3x^2y=0$ and $y_2$ solves $2y''+y'+3x^2y=e^x$. Which of the following is a solution of $2y''+y'+3x^2y=-2e^x$?

(A) $3y_1-2y_2$

(B) $y_1+2y_2$

(C) $2y_1-y_2$

(D) $y_1+2y_2-2e^x$

(E) None of the above.

The answer is supposed to be $A$. But I am not really sure how that happened.

I know that for $2y''+y'+3x^2y=-2e^x$ the solution is always the homogeneous part and the particular part added together.

Furthermore, I know that the homogeneous part is given as $y_1$.

I then know that for $2y''+y'+3x^2y=e^x$ the solution for that one is composed of the homogeneous part and the particular part and that I can also write the ODE as $-4y''-2y'-6x^2y=-2e^x$. So this implies that the homogeneous portion is just $-2y_1$.

I can't get further than that though. Is my thought process right so far? If not, what more can I do and how can I proceed from here? I can't even solve the first two equations since they have no analytic solution.

This is from an old exam I was looking at an not an assignment so feel free to show work.

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3 Answers 3

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You are on the right track. Just notice that $ay_1$ is a solution to the homogeneous for any constant a. So, when you say that "the solution to the homogeneous is just $-2y_1$", you could also say that $3y_1$ is also a solution. Then substitute $-2y_2$ in the LHS, and use the information you have about $y_2$, what do you get?

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  • $\begingroup$ Thanks a lot! I got it :) . $\endgroup$ Commented Jun 22, 2020 at 9:13
  • $\begingroup$ You're welcome! $\endgroup$ Commented Jun 22, 2020 at 22:56
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This is just linear algebra. On the left side you have a linear (differential, but that is not so important here) operator, call it $L$. Then you are given $L(y_1)=0$ and $L(y_2)=f$. Now you want to solve $L(y)=-2f$. By combining the known solutions you get $L(cy_1-2y_2)=-2f$ for any coefficient $c$.

Now it remains to check that $L(e^x)$ does not result in any usable right side.

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  • $\begingroup$ That's a very neat way of looking at this! $\endgroup$ Commented Jun 22, 2020 at 9:14
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By linearity, it suffices to substitute the RHS:

  • a) $3y_1-2y_2\to-2e^x,$

  • b) $y_1+2y_2\to2e^x,$

  • c) $2y_1-y_2\to-e^x,$

  • d) $y_1+y_2-2e^x\to -e^x.$

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  • $\begingroup$ Yes I just realized that now. Thank you :) . $\endgroup$ Commented Jun 22, 2020 at 9:15

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