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Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations2 Marks
Question
Verify that the function $y = x \sin x ($explicit or implicit$)$ is a solution of differential equation $xy' = y + x\sqrt {{x^2} - {y^2}} $ $\left( {x \ne 0\,\,and\,\,x > y\,or\,x < - y\,} \right)$
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Answer

Given: $y = x sin x …(i)$
To prove:y given by eq. $(i)$ is a solution of differential equation $xy' = y + x\sqrt {{x^2} - {y^2}} ...(ii)$
Proof: From eq. $(i),$
$\frac{{dy}}{{dx}}\left( { = y'} \right) = x\frac{d}{{dx}}\sin x + \sin x\frac{d}{{dx}}x$
$= x \cos x + \sin x$
$L.H.S.$ of eq. $(ii)$
$= xy' = x(x \cos x + \sin x) = x^2 \cos x + x \sin x$
$R.H.S.$ of eq. $(ii)$
$ = y + x\sqrt {{x^2} - {y^2}} $
$ = x\sin x + x\sqrt {{x^2} - {x^2}{{\sin }^2}x} [$from eq. $(i)]$
$ = x\sin x + x\sqrt {{x^2}\left( {1 - {{\sin }^2}x} \right)} $
$= x\sin x + x\sqrt {{x^2}{{\cos }^2}x} $
$= x\sin x + x.x\cos x$
$= x\sin x + {x^2}\cos x$
$= {x^2}\cos x + x\sin x$
$\therefore L.H.S. = R.H.S$
Hence, $y$ given by eq. $(i)$ is a solution of $xy' = y + x\sqrt {{x^2} - {y^2}} $.

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