Question

The equation y=xsinx and wants to prove x dpower 3y/dxpower3+x dy/dx+2y=0

Answer 1

$y=x\sin x$

$\dfrac{\mathrm dy}{\mathrm dx}=x\dfrac{\mathrm d(\sin x)}{\mathrm dx}+\dfrac{\mathrm dx}{\mathrm dx}\sin x=x\cos x+\sin x$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=x\dfrac{\mathrm d(\cos x)}{\mathrm dx}+\dfrac{\mathrm dx}{\mathrm dx}\cos x+\dfrac{\mathrm d(\sin x)}{\mathrm dx}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-x\sin x+\cos x+\cos x$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-x\sin x+2\cos x$

$\dfrac{\mathrm d^3y}{\mathrm dx^3}=-\left(x\dfrac{\mathrm d(\sin x)}{\mathrm dx}+\dfrac{\mathrm dx}{\mathrm dx}\sin x\right)+2\dfrac{\mathrm d(\cos x)}{\mathrm dx}$
$\dfrac{\mathrm d^3y}{\mathrm dx^3}=-x\cos x-\sin x-2\sin x$
$\dfrac{\mathrm d^3y}{\mathrm dx^3}=-x\cos x-3\sin x$

Substitute the derivatives into the given ODE and check to see if the relation is an identity.

$x\dfrac{\mathrm d^3y}{\mathrm dx^3}+x\dfrac{\mathrm dy}{\mathrm dx}+2y=0$
$x(-x\cos x-3\sin x)+x(x\cos x+\sin x)+2x\sin x=0$
$-x^2\cos x-3x\sin x+x^2\cos x+x\sin x+2x\sin x=0$
$0=0$

and we're done.