Question

Solve the differential equation: $\frac{dz}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2$

• A. $\frac{1}{-x\log z}=\frac{1}{2x^2}-c.$
• B. $\frac{1}{x\log z}=\frac{1}{2x^{-2}}-c.$
• C. $\frac{1}{x\log z}=\frac{-1}{2x^2}-c.$
• D. $\frac{1}{x\log z}=\frac{1}{2x^2}-c.$

Answer 1

The correct option is D $\frac{1}{x\log z}=\frac{1}{2x^2}-c.$

$\frac{dz}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2\Rightarrow \frac{1}{z{\left(\log z\right)}^2}\cdot \frac{dz}{dx}+\frac{1}{x}\cdot \frac{1}{\log z}=\frac{1}{x^2}$
Put $\frac{1}{\log z}=v\Rightarrow -\frac{1}{z{\left(\log z\right)}^2}dz=dv$
$\therefore \frac{dv}{dx}-\frac{1}{x}v=\frac{1}{x^2}$ ...(1)
Here $P=-\frac{1}{x}\Rightarrow \int Pdx=-\int \frac{1}{x}dx=-\log x=\log \frac{1}{x}$
$\therefore I.F.=e^{\log \frac{1}{x}}=\frac{1}{x}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{x}\frac{dv}{dx}-\frac{1}{x^2}v=\frac{1}{x^3}$
Integrating both sides we get
$\frac{v}{x}=\int \frac{1}{x^3}dx+c\Rightarrow -\frac{1}{2x^2}+c\Rightarrow \frac{1}{x\log z}=\frac{1}{2x^2}-c$