# Solution to a differential equation using equations that have no analytic solution

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.

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?

• Thanks a lot! I got it :) . Commented Jun 22, 2020 at 9:13
• You're welcome! Commented Jun 22, 2020 at 22:56

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.

• That's a very neat way of looking at this! Commented Jun 22, 2020 at 9:14

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) $$y_1+y_2-2e^x\to -e^x.$$

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