Which of the following transformation reduces the differential equation $\frac{dz}{dx}+\frac{z}{x}log z =\frac{z}{x^2} (log z)^2$ into the form $\frac{du}{dx}+ P (x) u = Q (x)$?
Answer & explanation
Correct answer: option 3
Dividing the given equation by z (log z)2, we get
$\frac{1}{z (log z)^2}\frac{dz}{dx}+\frac{1}{logz}\frac{1}{x}=\frac{1}{x^2}$ … (1)
Writing $\frac{1}{logz}= u$,
we have $\frac{du}{dx}= - (log z)^{-2}\frac{1}{z}\frac{dz}{dx}$
Hence (1) can be written as
$-\frac{du}{dx}+\frac{u}{x}=\frac{1}{x^2} ⇒\frac{du}{dx}-\frac{u}{x}=\frac{-1}{x^2}$
which is the required form with $P (x) =\frac{-1}{x}$ and $Q (x) =\frac{-1}{x^2}$
Hence (C) is the correct answer.