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D Y D X = ( X + Y + 1 ) 2

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Question

\[\frac{dy}{dx} = \left( x + y + 1 \right)^2\]
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Solution

We have, 

\[\frac{dy}{dx} = \left( x + y + 1 \right)^2 \]

\[\text{ Putting }x + y + 1 = v\]

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

\[ \therefore \frac{dv}{dx} - 1 = v^2 \]

\[ \Rightarrow \frac{dv}{dx} = v^2 + 1\]

\[ \Rightarrow \frac{1}{v^2 + 1}dv = dx\]

Integrating both sides, we get

\[\int\frac{1}{v^2 + 1}dv = \int dx\]

\[ \Rightarrow \tan^{- 1} v = x + C\]

\[ \Rightarrow \tan^{- 1} \left( x + y + 1 \right) = x + C\]

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Chapter 21: Differential Equations - Exercise 22.08 [Page 66]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 21 Differential Equations
Exercise 22.08 | Q 1 | Page 66

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