Question

Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?

Options

• x

• e^x

• log x

• log (log x)

Answer 1

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,

\[\frac{dy}{dx} + \frac{y}{x \log x} = 2\frac{\log x}{x \log x}\]

\[ \Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x}\]

\[ \Rightarrow \frac{dy}{dx} + \left( \frac{1}{x \log x} \right)y = \frac{2}{x}\]

\[\text{ Comparing with }\frac{dy}{dx} + Py = Q \text{ we get, }\]

\[P = \frac{1}{x \log x} \text{ and }Q = \frac{2}{x}\]
\[\text{ Now, }I . F = e^{\int P\ dx} = e^{\int\frac{1}{x\log x}dx} \]

\[ = e^{log\left( \log x \right)} \]

\[ = \log x\]