Question

The integrating factor of the differential equation (x log x) $\frac{dy}{dx}+y=2\log x$, is given by
(a) log (log x)

(b) e^x

(c) log x

(d) x

Answer 1

(c) log x

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