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I need a hand for solving the integration part of the differential equation $y''+4y=x^2sin2x$ . $(D-2i)(D+2i)y=x^2sin2x$ , $t= \dfrac{x{^2}sin2x}{D+2i}$

$t'+2it=x^2sin2x$, $t=uv$

$v=e^{-2ix}$

$du=(e^{2ix}) x^2sin2xdx$

I am stuck at this part. Can someone help me to solve it? Thanks in advance.

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    $\begingroup$ What do you want? Particular integration? $\endgroup$
    – Empty
    Nov 11, 2014 at 20:54
  • $\begingroup$ I want to find solution of the non-homogeneous part. For this, first I have to find u. $\endgroup$
    – Saruman
    Nov 11, 2014 at 21:00
  • $\begingroup$ I think It is called factorizing method in English. $\endgroup$
    – Saruman
    Nov 11, 2014 at 22:08
  • $\begingroup$ Can you show a few steps from the solution? $\endgroup$
    – Saruman
    Nov 11, 2014 at 22:54
  • $\begingroup$ Successive Integration of Two First-Order Equations. $\endgroup$
    – Saruman
    Nov 11, 2014 at 23:08

2 Answers 2

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Let, $P=\dfrac{1}{D^{2}+4}x^{2}\sin2x$ & $Q=\dfrac{1}{D^{2}+4}x^{2}\cos2x$.

Now, $Q+iP=\dfrac{1}{D^{2}+4}x^{2}e^{2ix}=e^{2ix}\dfrac{1}{D^{2}+4iD}x^{2}=e^{2ix}\dfrac{1}{4iD}(1+\dfrac{D}{4i})^{-1}x^{2}=\dfrac{e^{2ix}}{4i}(x^{3}-\dfrac{x^{2}}{4i}-\dfrac{x}{8})$.

Then, P.I.=$P$=imaginary part=$\dfrac{1}{16}x^{2}\sin2x-\dfrac{1}{4}(x^{3}-\dfrac{x}{8})\cos2x$.

If my calculation is right then it is the answer.But in calculation may be some mistake.Check It !!

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  • $\begingroup$ How did $D^2+4$ become $D^2+4iD$? $\endgroup$
    – Saruman
    Nov 11, 2014 at 21:46
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    $\begingroup$ Replacing $D$ by $(D+2i)$. First read your text book carefully. $\endgroup$
    – Empty
    Nov 12, 2014 at 3:25
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I don't know what approach you might be following here. I think you're trying to use variation of parameters. In which case

  • Find two independent solutions, $y_1$, $y_2$, to the homogeneous equation $y'' + 4y = 0$
  • Write the particular solution as $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$
  • Then look up (or re-derive!) how to write $u_1' = something$, $u_2' = other \ something$
  • Integrate to find $u_1$ and $u_2$, and hence $y_p$
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  • $\begingroup$ Actually, I know how to solve this kind of equations, but my problem is with the integration part. I want to find $u$ from this integral $du=(e^{2ix}) x^2sin2xdx$. $\endgroup$
    – Saruman
    Nov 11, 2014 at 21:49

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