Question

Find the solution of $\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=x^2+2y^2+\frac{y^4}{x^2}$

• A. $\frac{y}{x}-\frac{1}{2(x^2+y^2)}=c$
• B. $\frac{x}{y}-\frac{1}{2(x^2+y^2)}=c$
• C. $\frac{y}{x}+\frac{1}{2(x^2+y^2)}=c$
• D. $\frac{x}{y}-\frac{1}{2(x^2-y^2)}=c$

Answer 1

The correct option is A $\frac{y}{x}-\frac{1}{2(x^2+y^2)}=c$

Given, $\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=x^2+2y^2+\frac{y^4}{x^2}=\frac{x^4+2x^2{y}^2+y^4}{x^2}$
$\frac{xdx+ydy}{ydx-xdy}=\frac{{(x^2+y^2)}^2}{x^2}$
$\frac{xdx+ydy}{{(x^2+y^2)}^2}=\frac{-(xdy-ydx)}{x^2}$
$\frac{1}{2}\frac{2xdx+2ydy}{{(x^2+y^2)}^2}=-d\left(\frac{y}{x}\right)$
$\frac{1}{2}\frac{d(x^2+y^2)}{{(x^2+y^2)}^2}=-d\left(\frac{y}{x}\right)$
Now integrating both sides
$-\frac{1}{2}\frac{1}{x^2+y^2}=-\frac{y}{x}+c$
$\frac{y}{x}-\frac{1}{2(x^2+y^2)}=c$