Question

Solve:

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

• A. $xy=c(x-\sqrt{y^2-x^2})$
• B. $x=c(y+y^2-x^2)$
• C. $y=c(x+y^2-x^2)$
• D. $xy=c(y+\sqrt{y^2-x^2})$

Answer 1

The correct option is D $xy=c(y+\sqrt{y^2-x^2})$

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

$\Rightarrow \frac{dy}{dx}=\frac{y^3+y^2\sqrt{y^2-x^2}}{x^3}$
Put $\frac{y}{x}=v$
$\therefore v+x\frac{dv}{dx}=v^3+v^2\sqrt{v^2-1}$
$\Rightarrow \frac{dv}{v(v^2-1)+v^2\sqrt{v^2-1}}=\frac{dx}{x}$
$\Rightarrow \int \frac{dv}{v\sqrt{v^2-1}+v^2\sqrt{v^2-1}}=\int \frac{dx}{x}$
$\Rightarrow \log (v+\sqrt{v^2-1})-\log v=\log x+c$

$\Rightarrow \log \frac{v+\sqrt{v^2-1}}{v}=\log x-c$
$\therefore xy=c(y+\sqrt{y^2-x^2})$