Which of the following transformations reduce the differential equation $\frac{dz}{dx}+\frac{z}{x} logz=\frac{z}{x^2}(logz)^2$ into the form $\frac{du}{dx} + P(x) u = Q(x)$
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : $u=(log\, z)^{-1}$
We have,
$\frac{1}{z(log\, z)^2}\frac{dz}{dx}+\frac{1}{x}×\frac{1}{log\, z}=\frac{1}{x^2}$
Putting $\frac{1}{log \, z}=u $ and $-\frac{1}{z(log\, z)^2}\frac{dz}{dx}=\frac{du}{dx}, $ we get
$-\frac{du}{dx}+\frac{u}{x} =\frac{1}{x^2}$
$⇒\frac{du}{dx} +\left(-\frac{1}{x}\right) u = - \frac{1}{x^2}$ or, $\frac{du}{dx}+P(x) u = Q(x)$, where
$P(x) = - \frac{1}{x} $ and $Q(x) =-\frac{1}{x^2}$