Question

The solution of $\frac{x^{3}dx+y{x}^{2}dy}{\sqrt{x^2+y^2}}=ydx-xdy$ is:

• A. $\sqrt{x^2+y^2}=Cx$
• B. $\sqrt{x^2+y^2}+y/x=C$
• C. $\sqrt{x^2+y^2}+y/x^2=C$
• D. $(x^2+y^2{)}^2+x{y}^2=C$

Answer 1

The correct option is B $\sqrt{x^2+y^2}+y/x=C$

Given : $\frac{x^{3}dx+y{x}^{2}dy}{\sqrt{x^2+y^2}}=ydx-xdy$

$⟹\frac{x^2(xdx+ydy)}{\sqrt{x^2+y^2}}=ydx-xdy$

$⟹\frac{x^2(2xdx+2ydy)}{2\sqrt{x^2+y^2}}=-(xdy-ydx)$

$⟹\frac{x^2}{2}\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}=-x^{2}d\left(\frac{y}{x}\right)$ ......... [Using product rule fomula]

After integrating we get:
$\sqrt{x^2+y^2}=-y/x+C$

$⟹\sqrt{x^2+y^2}+y/x=C$

Hence, option B is correct.