Question
The solution of the differential equation $$\left\{ {1 + x\sqrt {\left( {{x^2} + {y^2}} \right)} } \right\}dx + \left\{ {\sqrt {\left( {{x^2} + {y^2}} \right)} - 1} \right\}y\,dy = 0{\text{ is :}}$$
A.
$${x^2} + \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C$$
B.
$$x - \frac{{{y^2}}}{3} + \frac{1}{2}{\left( {{x^2} + {y^2}} \right)^{\frac{1}{2}}} = C$$
C.
$$x - \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C$$
D.
none of these
Answer :
$$x - \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C$$
Solution :
Rearranging the equation, we have
$$\eqalign{
& dx - y\,dy + \sqrt {\left( {{x^2} + {y^2}} \right)} \left( {x\,dx + y\,dy} \right) = 0 \cr
& \Rightarrow dx - y\,dy + \frac{1}{2}\sqrt {\left( {{x^2} + {y^2}} \right)} d\left( {{x^2} + {y^2}} \right) = 0 \cr} $$
On integrating, we get
$$\eqalign{
& x - \frac{{{y^2}}}{2} + \frac{1}{2}\int {\sqrt t \,dt = c,\,\left\{ {t = \sqrt {\left( {{x^2} + {y^2}} \right)} } \right\}} \cr
& {\text{or }}x - \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C \cr} $$