Question

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

Options

• log (log x)

• e^x

• log x

• x

Answer 1

log x

 

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