Question
Verify that the function $y = \sqrt {{a^2} - {x^2}} , $ $x\in \left( { - a,a} \right)$ (explicit or implicit) is a solution of differential equation  $x + y\frac{{dy}}{{dx}} = 0\left( {y \ne 0} \right)$

Answer

Given: $y = \sqrt {{a^2} - {x^2}} , x\in \left( { - a,a} \right)$ …(i)
To prove:y given by eq. (i) is a solution of differential equation $x + y\frac{{dy}}{{dx}} = 0$ …(ii)
Proof: From eq. (i),
$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{a^2} - {x^2}} }}\left( { - 2x} \right)$
$ = \frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}$ …(iii)
Putting the values of y and $\frac{{dy}}{{dx}}$ from eq. (i) and (iii) in L.H.S. of eq. (ii),
$ = x + y\frac{{dy}}{{dx}}$
$= x + \sqrt {{a^2} - {x^2}} \left( {\frac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}} \right)$
= x - x = 0 = R.H.S. of eq. (ii)
Hence, function given by eq. (i) is a solution of $x + y\frac{{dy}}{{dx}} = 0$.

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