Find a second-order homogeneous linear ODE for which the given functions are solutions: 1, e^(-2x)?

I have come across this question in my textbook, but have no idea how to solve it. The basis of this wasn't covered in the book. Up to this point in the book, I've just been solving many differential equation. However, this seems to be wanting to work backwards.

I was wondering if anybody knew how to solve it?

The rest of the question is with regard to solving the equation, which I can do simply.

The help is much appreciated.

  • $\begingroup$ Hint: $y'' - 2 y' = 0$. We would solve $m(m-2) = 0$. Now find the initial conditions to make it work. $\endgroup$ – Moo Sep 8 '16 at 3:04
  • $\begingroup$ I see, so that would be to find a particular solution? Would the ODE itself just be y''-2y'? $\endgroup$ – M.MacKinnon Sep 8 '16 at 3:23
  • $\begingroup$ Yes, that is the ODE. The solution to that is $y(x) = c_1 e^{-2x} + c_2$. Now, you need initial conditions such that $c_1 = c_2 = 1$. $\endgroup$ – Moo Sep 8 '16 at 3:30
  • $\begingroup$ I found the initial conditions y(0)=1 and y'(0)=-1 to work. Thanks a lot for your help. $\endgroup$ – M.MacKinnon Sep 8 '16 at 3:56

Hint: $y'' + b y' + c y = 0$ has $e^{r t}$ as a solution if $r$ is a root of the quadratic $r^2 + b r + c = 0$. What quadratic has roots $0$ and $-2$?

  • $\begingroup$ r^(2)+2r has those roots.Where would you proceed from there? $\endgroup$ – M.MacKinnon Sep 8 '16 at 3:13
  • $\begingroup$ Thanks a lot for your help, Robert. $\endgroup$ – M.MacKinnon Sep 8 '16 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.