# Solve the differential equation: (1 + y2) dx = (tan−1 y − x) dy - Mathematics and Statistics

Sum

Solve the differential equation:

(1 + y2) dx = (tan1 y x) dy

#### Solution

We have,

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

$\Rightarrow \frac{dx}{dy} = \frac{\tan^{- 1} y - x}{1 + y^2}$

$\Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{- 1} y}{1 + y^2}$

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

$P = \frac{1}{1 + y^2}$

$Q = \frac{\tan^{- 1} y}{1 + y^2}$

Now,

$I . F . = e^{\int\frac{1}{1 + y^2}dy} = e^{\tan^{- 1} y}$

So, the solution is given by

$x \times e^{\tan^{- 1} y} = \int\frac{\tan^{- 1} y}{1 + y^2} \times e^{\tan^{- 1} y} dy + C$

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

Now,

$I = \int\frac{\tan^{- 1} y}{1 + y^2} \times e^{\tan^{- 1} y} dy$

$\text{Putting }t = \tan^{- 1} y,\text{ we get}$

$dt = \frac{1}{1 + y^2}dy$

$\therefore I = \int\underset{I}{t}\times \underset{II}{e^t} \text{ }dt$

$= t \times \int e^t dt - \int\left( \frac{d t}{d t} \times \int e^t dt \right)dt$

$= t e^t - \int e^t dt$

$= t e^t - e^t$

$\therefore I = \tan^{- 1} y e^{\tan^{- 1} y} - e^{\tan^{- 1} y}$

$= e^{\tan^{- 1} y} \left( \tan^{- 1} y - 1 \right)$

Putting the value of I in (1), we get

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

Is there an error in this question or solution?
Chapter 22: Differential Equations - Revision Exercise [Page 146]

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 60 | Page 146
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