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

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Solution

Let $y = x sin x$
Differentiating both sides w.r.t. $x$ we get,
$y^{'} = (x sin x)^{'}$
$y^{'} = (x)^{'} sin x + x . (sin x)^{'}$
$y^{'} = 1. sin x + x (cos x)$
$y^{'} = sin x + x cos x$
Taking LHS
Putting
$y^{'} = sin x + x cos x$
$x y^{'} = x (sin x + x cos x)$
$x y^{'} = x sin x + x^{2} cos x$
Taking RHS
Putting $y = x sin x$
$y + x \sqrt{x^{2} - y^{2}} = x sin x + x \sqrt{x^{2} - (x sin x)^{2}}$
$y + x \sqrt{x^{2} - y^{2}} = x sin x + x \sqrt{x^{2} (1 - {sin}^{2} x)} (1 - {sin}^{2} x = {cos}^{2} x)$
$y + x \sqrt{x^{2} - y^{2}} = x sin x + x^{2} \sqrt{c o s^{2} x}$
$y + x \sqrt{x^{2} - y^{2}} = x sin x + x^{2} cos x$
Thus $L H S = R H S$
Hence verified.

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

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