# (X2 + 1) Dy + (2y − 1) Dx = 0 - Mathematics

Sum

(x2 + 1) dy + (2y − 1) dx = 0

#### Solution

We have,

$\left( 1 + x^2 \right)dy + \left( 2y - 1 \right)dx = 0$

$\Rightarrow \left( 1 + x^2 \right)dy = \left( 1 - 2y \right)dx$

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

Integrating both sides, we get

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

$\Rightarrow - \frac{1}{2}\log\left| 1 - 2y \right| = \tan^{- 1} x - \log \sqrt{C}$

$\Rightarrow - \log\left| 1 - 2y \right| = 2 \tan^{- 1} x - 2\log \sqrt{C}$

$\Rightarrow - 2 \tan^{- 1} x = - \log C + \log\left| 1 - 2y \right|$

$\Rightarrow - 2 \tan^{- 1} x = \log \left| \frac{1 - 2y}{C} \right|$

$\Rightarrow e^{- 2 \tan^{- 1} x} = \frac{1 - 2y}{C}$

$\Rightarrow C e^{- 2 \tan^{- 1} x} = \left( 1 - 2y \right)$

$\Rightarrow 1 - C e^{- 2 \tan^{- 1} x} = 2y$

$\Rightarrow \frac{1}{2} - \frac{C}{2} e^{- 2 \tan^{- 1} x} = y$

$\Rightarrow y = \frac{1}{2} + K e^{- 2 \tan^{- 1} x},\text{ where }K = - \frac{C}{2}$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 45 | Page 146