Question

The solution of the differential equation
$\frac{dy}{dx}+\frac{y}{x}logy=\frac{y}{x^2}{\left(logy\right)}^2$ is

• A. $y=log\left(x^2+cx\right)$
• B. $logy=x\left(c{x}^2+\frac{1}{2}\right)$
• C. $x=logy\left(c{x}^2+\frac{1}{2}\right)$
• D. None of these

Answer 1

Answer: (C)

Divide the equation by $y(logy{)}^2$
$\frac{1}{y{\left(logy\right)}^2}\frac{dy}{dx}+\frac{1}{logy}\cdot \frac{1}{x}=\frac{1}{x^2}$
Put $\frac{1}{logy}=z\Rightarrow \frac{-1}{y{\left(logy\right)}^2}\frac{dy}{dx}=\frac{dz}{dx}$
Thus, we get, $-\frac{dz}{dx}+\frac{1}{x}\cdot z=\frac{1}{x^2}$, linear in z
$\Rightarrow \frac{dz}{dx}+\left(-\frac{1}{x}\right)z=-\frac{1}{x^2}$
I.F. $=e^{-\int \frac{1}{x}dx}=e^{-logx}=\frac{1}{x}$
$\therefore$ The solution is, $z\left(\frac{1}{x}\right)=\int \frac{-1}{x^2}\left(\frac{1}{x}\right)dx+c$
$\Rightarrow \frac{1}{logy}\left(\frac{1}{x}\right)=\frac{-x^{-2}}{-2}+c\Rightarrow xlogy\left(c{x}^2+\frac{1}{2}\right)$