# Solve the Following Differential Equation:- (1 + X2) Dy + 2xy Dx = Cot X Dx - Mathematics

Sum

Solve the following differential equation:-

(1 + x2) dy + 2xy dx = cot x dx

#### Solution

We have,

$\left( 1 + x^2 \right)dy + 2xy dx = \cot x dx$

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

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

$P = \frac{2x}{\left( 1 + x^2 \right)}$

$Q = \frac{\cot x}{\left( 1 + x^2 \right)}$

Now,

$I.F. = e^{\int\frac{2x}{\left( 1 + x^2 \right)}dx}$

$= e^{\log\left| 1 + x^2 \right|}$

$= 1 + x^2$

So, the solution is given by

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

$\Rightarrow y\left( 1 + x^2 \right) = \int\left[ \frac{\cot x}{\left( 1 + x^2 \right)} \times \left( 1 + x^2 \right) \right] dx + C$

$\Rightarrow y\left( 1 + x^2 \right) = \int\cot x dx + C$

$\Rightarrow y\left( 1 + x^2 \right) = \log \left| \sin x \right| + C$

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 66.12 | Page 147