Question

The solution of the differential equation
$\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0$ is

• A. $x^{2}+\frac{y^{2}}{2}+\frac{1}{3}\left(x^{2}+y^{2}\right)^{3 / 2}=c$
• B. $x-\frac{y^{3}}{3}+\frac{1}{2}\left(x^{2}+y^{2}\right)^{1 / 2}=c$
• C. $x-\frac{y^{2}}{2}+\frac{1}{3}\left(x^{2}+y^{2}\right)^{3 / 2}=c$
• D. None of these

Answer 1

Answer: (C)

Rearranging the equation, we have
$ d x-y\,d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y \,d y)=0 $
$\Rightarrow d x-y \,d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)} d\left(x^{2}+y^{2}\right)=0$
On integrating, we get
$x-\frac{y^{2}}{2}+\frac{1}{2} \int \sqrt{t} \,d t=c,\left\{t=\sqrt{\left(x^{2}+y^{2}\right)}\right\}$
or $x-\frac{y^{2}}{2}+\frac{1}{3}\left(x^{2}+y^{2}\right)^{3 / 2}=c$