$y = x\sin x:xy' = y + x\sqrt {{x^2} - {y^2}} (x \ne 0$ and $x > y$or $x < - y)$

.: We have, $y = x\sin x$

…(1)
Differentiating (1) w.r.t. $x$,

we get $y' = x\cos x + \sin x$

…(2)
Multiplying (2) by $x$,

we get
$xy' = {x^2}\cos x + x\sin x$
$\Rightarrow xy' = {x^2}\sqrt {1 - {{\sin }^2}x} + y$
[from (1)]
$\Rightarrow xy' = {x^2}\sqrt {1 - {{\left( {\cfrac{y}{x}} \right)}^2}} + y$

(From (1) $\sin x = \cfrac{y}{x}$ )
$\Rightarrow xy' = x\sqrt {{x^2} - {y^2}} + y$

$\therefore$ $y = x\sin x$ is a solution of
the given differential equation.