Question

Which of the following is the integrating factor of (x log x) $\frac{dy}{dx}+y$ = 2 log x?
(a) x
(b) e^x
(c) log x
(d) log (log x)

Answer 1

(c) log x

We have,
$\left(x\log x\right)\frac{dy}{dx}+y=2\log x$
Dividing both sides by (x log x) we get,
$$\begin{gathered} \frac{dy}{dx}+\frac{y}{x\log x}=2\frac{\log x}{x\log x} \\ \Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x} \\ \Rightarrow \frac{dy}{dx}+\left(\frac{1}{x\log x}\right)y=\frac{2}{x} \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q\mathrm{we}\mathrm{get}, \\ P=\frac{1}{x\log x}\mathrm{and}Q=\frac{2}{x} \\ \mathrm{Now},\mathrm{I}.\mathrm{F}=e^{\int Pdx}=e^{\int \frac{1}{x\log x}dx} \\ =e^{\log \left(\log x\right)} \\ =\log x \end{gathered}$$