osnomu3

2021-12-27

$yy=2x\mathrm{sin}x$

jgardner33v4

The differential equation shown is
$yy=2x\mathrm{sin}x$
$y=A{e}^{mx}\left(A\ne 0\right)$ be the trail solution of the given differential equation
The auxilary equation follows is
${m}^{2}+1=0$
$y{m}^{2}=-1$
$ym=1i$
$cf={c}_{1}x+{c}_{2}\mathrm{sin}x$
$PI=\frac{1}{{D}^{2}+1}\left(2x\mathrm{sin}x\right)$
$=2\frac{1}{{D}^{2}+1}\left(x\mathrm{sin}x\right)$
$=2\text{imarginary part of}\frac{1}{{D}^{2}+1}x.{e}^{ix}$
$=2.I.P.O{e}^{ix}\frac{1}{{\left(D+1\right)}^{2}+1}x$
$=2IPO{e}^{ix}\frac{1}{{D}^{2}+2D+2}x$
$=2IPO{e}^{ix}\frac{1}{2\left(1+\frac{{D}^{2}+2D}{2}\right)}x$
$=IPO{e}^{ix}{\left(1+\frac{{D}^{2}+2D}{2}\right)}^{-1}x$
$=IPO{e}^{ix}\left(1-\frac{{D}^{2}+2D}{2}+\dots \right)x$
$IPO{e}^{ix}\left(x-\frac{2}{2}\right)$
$=IPO{e}^{ix}\left(x-1\right)$
$=IPO\left({c}_{3}x+i\mathrm{sin}x\right)\left(x-1\right)$
$=x\mathrm{sin}x-\mathrm{sin}x$
Therefore $y\left(x\right)=cf+\pi$
$={c}_{1}{c}_{3}x+{c}_{2}\mathrm{sin}x+x\mathrm{sin}x-\mathrm{sin}x$
$=\left({c}_{2}+x-1\right)\mathrm{sin}x+{c}_{1}{c}_{3}x$

Neunassauk8

$yy=2x\mathrm{sin}x$
$y=\left(ax+b\right)\mathrm{cos}x+\left(cx+d\right)\mathrm{sin}x$
${y}^{\prime }=a\mathrm{cos}x+c\mathrm{sin}x+\left(ax+b\right)\left(-\mathrm{sin}x\right)+\left(cx+d\right)\mathrm{cos}x$
$y-2a\mathrm{sin}x+2\mathrm{cos}x-\left(ax+b\right)\mathrm{cos}x+\left(cx+d\right)\left(-\mathrm{sin}x\right)$
$yy=-2a\mathrm{sin}x+2\mathrm{cos}x$
x turn is not including
our choice is not corvet
Take $y=\left(a{x}^{2}+bx+c\right)\mathrm{cos}x+\left(b{x}^{2}+ex+f\right)\mathrm{sin}x$
${y}^{\prime }=\left(2ax+b\right)\mathrm{cos}v+\left(2dx+e\right)\mathrm{sin}x+\left(a{x}^{2}+bx+c\right)\left(-\mathrm{sin}x\right)+\left({dx}^{2}+ex+f\right)\left(\mathrm{cos}x\right)$
$y2a\mathrm{cos}x+2d\mathrm{sin}x+\left(2ax+b\right)\left(-\mathrm{sin}x\right)+\left(2dx+e\right)\mathrm{cos}x+\left(2ax+b\right)\left(-\mathrm{sin}x\right)+\left(2dx+e\right)\mathrm{cos}x-\left(a{x}^{2}+bx+c\right)\mathrm{cos}x-\left({dx}^{2}+ex+f\right)\mathrm{cos}x$
$yy=2x\mathrm{sin}x$
$⇒\left(2a+2dx+e+2dx+e\right)\mathrm{cos}x+\left(2d-2ax-b-2ax-b\right)\mathrm{sin}x=2x\mathrm{sin}x$
$2a+4dx+2e=0⇒d=0;a+e=0$

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