Question

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

• A. $x^2+\frac{y^2}{2}+\frac{1}{3}{\left(x^2+y^2\right)}^{3/2}=C$
• B. $x-\frac{y^2}{3}+\frac{1}{2}{\left(x^2+y^2\right)}^{1/2}=C$
• 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 ha
$dx-ydy+\sqrt{\left(x^2+y^2\right)}\left(xdx+ydy\right)=0$
$\Rightarrow dx-ydy+\frac{1}{2}\sqrt{\left(x^2+y^2\right)}d\left(x^2+y^2\right)=0$
On integrating, we get
$x-\frac{y^2}{2}+\frac{1}{2}\int \sqrt{t}dt=C,\left\{t=\sqrt{\left(x^2+y^2\right)}\right\}$
or $x-\frac{y^2}{2}+\frac{1}{2}{\left(x^2+y^2\right)}^{3/2}=C$