Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y=x\sin x:x{y}^{\prime }=y+x\sqrt{x^2-y^2}(x\ne 0\text{and}x>y\text{or}x<-y)$

Answer 1

Let $y=x\sin x$
Differentiating both sides w.r.t. $x$ we get,
$y^{\prime }=(x\sin x{)}^{\prime }$
$y^{\prime }=(x{)}^{\prime }\sin x+x.(\sin x{)}^{\prime }$
$y^{\prime }=1.\sin x+x\left(\cos x\right)$
$y^{\prime }=\sin x+x\cos x$
Taking LHS
Putting
$y^{\prime }=\sin x+x\cos x$
$x{y}^{\prime }=x(\sin x+x\cos x)$
$x{y}^{\prime }=x\sin x+x^2\cos x$
Taking RHS
Putting $y=x\sin x$
$y+x\sqrt{x^2-y^2}=x\sin x+x\sqrt{x^2-(x\sin x{)}^2}$
$y+x\sqrt{x^2-y^2}=x\sin x+x\sqrt{x^2(1-{\sin }^{2}x)}(1-{\sin }^{2}x={\cos }^{2}x)$
$y+x\sqrt{x^2-y^2}=x\sin x+x^2\sqrt{co{s}^{2}x}$
$y+x\sqrt{x^2-y^2}=x\sin x+x^2\cos x$
Thus $LHS=RHS$
Hence verified.

Final answer:
Hence, the function $y=x\sin x$ is a solutio of the differential equation $x{y}^{\prime }=y+x\sqrt{x^2-y^2}$