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The solution of $y' = {(x + y)}^{2}, y (0) = 0$ is given by

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$\tan x = x + y$

$\sin^{2} y = x + y$

$\tan y = x + y$

$\sin y = x + y$

answer is A.

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Detailed Solution

Put $x + y = u \Rightarrow 1 + y^{'} = \frac{d u}{d x}$ The equation reduces to

$\frac{d u}{d x} - 1 = u^{2} \Rightarrow \frac{d u}{1 + u^{2}} = x$

$\Rightarrow \tan^{- 1} (x + y) = x + C$

But $y (0) = 0$ so $C = 0$

$\Rightarrow x + y = \tan x$

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