Question

Which of the following transformation reduces the differential equation $\frac{dz}{dx}+\frac{z}{x}\ln z=\frac{z}{x^2}(\ln z{)}^2$ into the form $\frac{du}{dx}+uP\left(x\right)=Q\left(x\right)$

• A. $y=\ln z$
• B. $u=e^z$
• C. $u=(\ln z{)}^{-1}$
• D. $u=(\ln z{)}^2$

Answer 1

The correct option is C $u=(\ln z{)}^{-1}$

Given :
$\frac{dz}{dx}+\frac{z}{x}\ln z=\frac{z}{x^2}(\ln z{)}^2$
$\Rightarrow \frac{1}{z(\ln z{)}^2}\frac{dz}{dx}+\frac{1}{x\ln z}=\frac{1}{x^2}$
Put
$\frac{1}{\ln z}=y\Rightarrow -\frac{1}{z(\ln z{)}^2}\frac{dz}{dx}=\frac{dy}{dx}$
So, Equation becomes,
$\frac{dy}{dx}-\frac{y}{x}=\frac{1}{x^2}$