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Question
\[\frac{dy}{dx} = \left( x + y \right)^2\]
Solution
We have,
\[\frac{dy}{dx} = \left( x + y \right)^2 . . . . . \left( 1 \right)\]
Let `x + y = v`
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
Therefore, (1) becomes
\[ \therefore \frac{dv}{dx} - 1 = v^2 \]
\[ \Rightarrow \frac{dv}{dx} = v^2 + 1\]
\[ \Rightarrow \frac{1}{v^2 + 1}dv = dx\]
Integrating both sides, we get
\[\int\frac{1}{v^2 + 1}dv = \int dx\]
\[ \Rightarrow \tan^{- 1} v = x + C\]
\[ \Rightarrow v = \tan\left( x + C \right)\]
\[ \Rightarrow x + y = \tan\left( x + C \right) \]
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