Mathematics - Differential Equation Question with Solution | TestHub

Given equ. Can be written as

$\begin{array}{rl}& \frac{dz}{z(\ln z{)}^2}+\frac{dx}{x(\ln z)}=\frac{dx}{x^2}\\ & Putu=\frac{1}{\ln z}\\ & \frac{du}{dx}=-\frac{1}{(\ln z{)}^2}\cdot \frac{1}{z}\cdot \frac{dz}{dx}\\ & \therefore -\frac{du}{dx}+\frac{1}{x}\cdot u=\frac{1}{x^2}\\ & \frac{du}{dx}+\left(-\frac{1}{x}\right)u=\left(-\frac{1}{x^2}\right)\\ & I.F.=e^{\int -\frac{1}{x}dx}=e^{\ln \frac{1}{x}}=\frac{1}{x}\\ & u\cdot \frac{1}{x}=\int -\frac{1}{x^2}\cdot \frac{1}{x}dx\\ & \frac{u}{x}=-\int \frac{1}{x^3}dx=+\frac{1}{2x^2}+c\end{array}$