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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.

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  • $\begingroup$ 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
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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

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

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  • $\begingroup$ 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

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