Question

The solution of differential equation $\frac{x+y\frac{dy}{dx}}{y-x\frac{dy}{dx}}=\frac{x{\cos }^2(x^2+y^2)}{y^3}$ is

• A. $\tan (x^2+y^2)=\frac{x^2}{y^2}+C$
• B. $\cot (x^2+y^2)=\frac{x^2}{y^2}+C$
• C. $\tan (x^2+y^2)=\frac{y^2}{x^2}+C$
• D. $\cot (x^2+y^2)=\frac{y^2}{x^2}+C$

Answer 1

The correct option is A

$\tan (x^2+y^2)=\frac{x^2}{y^2}+C$

$\frac{xdx+ydy}{y^2\left(\frac{ydx-xdy}{y^2}\right)}=\frac{xco{s}^2(x^2+y^2)}{y^3}$

$=\frac{1}{2}\int \frac{d(x^2+y^2)}{co{s}^2(x^2+y^2)}=\int \frac{x}{y}d\left(\frac{x}{y}\right)$

$=\frac{1}{2}\tan (x^2+y^2)=\frac{{\left(\frac{x}{y}\right)}^2}{2}+c$

$=\tan (x^2+y^2)=\frac{x^2}{y^2}+C$