Question

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

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

Answer 1

Answer: (C)

Divide the equation by $y(log\,y)^2$
$\frac{1}{y\left(log\,y\right)^{2}} \frac{dy}{dx}+\frac{1}{log\,y} \cdot \frac{1}{x} = \frac{1}{x^{2}}$
Put $\frac{1}{log\,y} = z \Rightarrow \frac{-1}{y\left(log\,y\right)^{2}} \frac{dy}{dx} = \frac{dz}{dx}$
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. $= e^{-\int \frac{1}{x}dx} = e ^{-log\,x} = \frac{1}{x}$
$\therefore $ The solution is, $z\left(\frac{1}{x}\right) = \int\frac{-1}{x^{2}}\left(\frac{1}{x}\right)dx + c$
$\Rightarrow \frac{1}{log\,y} \left(\frac{1}{x}\right) = \frac{-x^{-2}}{-2}+c\,\Rightarrow \,x log\, y\left(cx^{2} +\frac{1}{2}\right)$