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Differential Equations — Maths STD 12 Science — Question

Gujarat 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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