Question

Which of the following transformation reduce the differential equation
$\frac{dz}{dx}+\frac{z}{x}\log z=z\left(\log z{)}^{2}f\right(x)$ into the form
$\frac{du}{dx}+P\left(x\right)\left(u\right)=Q\left(x\right)$.

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

Answer 1

Answer: (C)

$u=(\log z{)}^{-1}$

$\frac{dz}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2$

Dividing the given equation by $z{\left(\log z\right)}^2$, we get

$\frac{1}{z{\left(\log z\right)}^2}\frac{dz}{dx}+\frac{1}{\log z}.\frac{1}{x}=\frac{1}{x^2}$

Putting $\frac{1}{\log z}=v\Rightarrow \frac{dv}{dx}=-{\left(\log z\right)}^{-2}\frac{1}{z}\frac{dz}{dx}$

$\therefore -\frac{dv}{dx}+\frac{v}{x}=\frac{1}{x^2}\Rightarrow \frac{dv}{dx}-\frac{v}{x}=-\frac{1}{x^2}$

which is in the required form with $P\left(x\right)=\frac{-1}{x}$ and $Q\left(x\right)=-\frac{1}{x^2}$