Question

Which of the following transformations reduce the differential equation $\frac{dz}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2$ into the form $\frac{du}{dx}+P\left(x\right)u=Q\left(x\right)$
(a) u = log x
(b) u = e^z
(c) u = (log z)⁻¹
(d) u = (log z)²

Answer 1

(c) u = (log z)⁻¹

$$\begin{gathered} \mathrm{Given}\frac{dz}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2.....\left(1\right) \\ \mathrm{Let}u={\left(\log z\right)}^{-1} \\ \frac{du}{dx}=-\frac{1}{{\left(\log z\right)}^2}\times \frac{1}{z}\times \frac{dz}{dx} \\ \frac{dz}{dx}=-z{\left(\log z\right)}^2\frac{du}{dx} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\frac{dz}{dx}\mathrm{from}\mathrm{equation}\left(1\right)\mathrm{we}\mathrm{get}, \\ \therefore -z{\left(\log z\right)}^2\frac{du}{dx}+\frac{z}{x}\log z=\frac{z}{x^2}{\left(\log z\right)}^2 \\ \frac{du}{dx}-\frac{1}{x}\frac{1}{\log z}=-\frac{1}{x^2} \\ \frac{du}{dx}-\frac{1}{x}{\left(\log z\right)}^{-1}=-\frac{1}{x^2} \\ \frac{du}{dx}-\frac{1}{x}u=-\frac{1}{x^2} \\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}, \\ \frac{du}{dx}+p\left(x\right)u=Q\left(x\right) \\ w\mathrm{here},p\left(x\right)=-\frac{1}{x} \\ q\left(x\right)=-\frac{1}{x^2} \end{gathered}$$


The correct option is C.