I have a differential equation as follows:

$$y''-4y'-4sy=\frac{4}{s} \mathrm{exp}(2x-2x\sqrt{1+s})$$

The general solution of the above equation would be in the form of

$$y = C_1 \exp{[(2+2\sqrt{1+s})x]}+C_2\, \exp{[(2-2\sqrt{1+s})x]}$$

I would appreciate any suggestions on solving for the particular solution.

Where s can be treated as a constant positive number. Thank you.

  • $\begingroup$ Method of undetermined coefficients. Look for $Dxe^{ax}$ with $a=2-2\sqrt{1+s}$. Note your homogeneous solution has a typo. You miss an $x$ after each first $2$ in the exponent. $\endgroup$ – Julien Apr 3 '13 at 17:18
  • $\begingroup$ @julien Yes, you are correct. I am missing x. $\endgroup$ – Jdbaba Apr 3 '13 at 17:46

The problem is that the RHS is already a solution to the differential equation. One way to deal with this is to use

$$y_p = A x \exp{[(2-\sqrt{1+s}) x]}$$

as a particular solution. When you plug into the diff eq'n, you get

$$A [2 (2-\sqrt{1+s}) - 4] = \frac{4}{s} \implies A = -\frac{1}{s \sqrt{1+s}}$$

  • $\begingroup$ Given the rhs, woudn't it be better to have a $+$ instead of the first $-$ in the exponent? $\endgroup$ – Julien Apr 3 '13 at 17:20
  • $\begingroup$ @julien: yep, it would. $\endgroup$ – Ron Gordon Apr 3 '13 at 17:24
  • $\begingroup$ @RonGordon When you assume yp , is it yp = A * exp(...) or yp = Ax exp(...). I think that should be a multiplier sign. What do you think ? $\endgroup$ – Jdbaba Apr 3 '13 at 17:53
  • $\begingroup$ @Jdbaba: no, it is the variable $x$. Note that this symbol is also used inside the exponential, so there is no ambiguity. $\endgroup$ – Ron Gordon Apr 3 '13 at 18:03
  • $\begingroup$ @ Ron If you look at the solution from Wolfram bit.ly/XqjYWs , the solution you provided is different from the solution given there. Would you please check this link ? $\endgroup$ – Jdbaba Apr 3 '13 at 20:32

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.