# Solve the Following Differential Equation:- X Log X D Y D X + Y = 2 X Log X - Mathematics

Sum

Solve the following differential equation:-

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

#### Solution

We have,

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

Dividing both sides by x log x, we get

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

$\Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^2}$

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

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

$P = \frac{1}{x \log x}$

$Q = \frac{2}{x^2}$

Now,

$I . F . = e^{\int P\ dx}$

$= e^{\int\frac{1}{x \log x}dx}$

$= e^{\log\left| \left( \log x \right) \right|}$

$= \log x$

So, the solution is given by

$y \times I . F . = \int Q \times I . F . dx + C$

$\Rightarrow y\log x = 2\int\frac{1}{x^2} \times \log x dx + C$

$\Rightarrow y\log x = I + C . . . . . . . . \left( 1 \right)$

Where,

$\Rightarrow I = 2\log x\int\frac{1}{x^2} dx - 2\int\left[ \frac{d}{dx}\left( \log x \right)\int\frac{1}{x^2} dx \right]dx$

$\Rightarrow I = \frac{- 2}{x}\log x + 2\int\left[ \frac{1}{x^2} \right]dx$

$\Rightarrow I = \frac{- 2}{x}\log x - \frac{2}{x} . . . . . . . . \left( 2 \right)$

From (1) and (2) we get

$\therefore y\log x = \frac{- 2}{x}\log x - \frac{2}{x} + C$

$\Rightarrow y\log x = \frac{- 2}{x}\left( \log x + 1 \right) + C$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 66.11 | Page 147