The solution of the differential equation dy dx = xy x^2 + y^2 is - Vidyadip

MCQ

The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}}$ is

✓
$a{y^2} = {e^{{x^2}/{y^2}}}$
B
$ay = {e^{x/y}}$
C
$y = {e^{{x^2}}} + {e^{{y^2}}} + c$
D
$y = {e^{{x^2}}} + {y^2} + c$

✓

Answer

Correct option: A.

$a{y^2} = {e^{{x^2}/{y^2}}}$

a
(a) Given $\frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}}$. Put $y = vx$; $\frac{{dy}}{{dx}} = v + x.\frac{{dv}}{{dx}}$

$\therefore v + x\frac{{dv}}{{dx}} = \frac{{(x)(vx)}}{{{x^2} + {v^2}{x^2}}}$

==> $v + x.\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}}$ ==> $x\frac{{dv}}{{dx}} = \frac{{ - {v^3}}}{{1 + {v^2}}}$

==> $\frac{{(1 + {v^2})}}{{{v^3}}}dv = - \frac{{dx}}{x}$ ==> $\left( {\frac{1}{{{v^3}}} + \frac{1}{v}} \right)dv = - \frac{{dx}}{x}$

Integrating both sides, $\int_{}^{} {\frac{{dv}}{{{v^3}}}} + \int_{}^{} {\frac{{dv}}{v}} = - \int_{}^{} {\frac{{dx}}{x}} $

==> $ - \frac{1}{{2{v^2}}} + \log v = - \log x - \log c$

==> $ - \frac{{{x^2}}}{{2{y^2}}} + \log y = - \log c$ ==> $\log cy = \frac{{{x^2}}}{{2{y^2}}}$

==> $cy = {e^{{x^2}/2{y^2}}}$ ==> ${c^2}{y^2} = {e^{{x^2}/{y^2}}}$

$\therefore {y^2}a = {e^{{x^2}/{y^2}}}$, where ${c^2} = a$.

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