Question

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

• A. $v=logz$
• B. $v=e^2$
• C. $v=\frac{1}{logz}$
• D. $v=(logz{)}^2$

Answer 1

Answer: (C)

$v=\frac{1}{logz}$

Given,

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

divide both side by $z(logz{)}^2$

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

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

Let $v=\frac{1}{logz}$ ..............1

and $\frac{dv}{dx}=\frac{-1}{z(logz{)}^2}\frac{dz}{dx}$ ................2

So , by 1 & 2

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

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

so, here $v=\frac{1}{logz}$