The solution of the equation dy dx = (x + y)^2 is - Vidyadip

MCQ

The solution of the equation $\frac{{dy}}{{dx}} = {(x + y)^2}$ is

A
$x + y + \tan (x + c) = 0$
B
$x - y + \tan (x + c) = 0$
✓
$x + y - \tan (x + c) = 0$
D
None of these

✓

Answer

Correct option: C.

$x + y - \tan (x + c) = 0$

c
(c) Put $x + y = v$ and $1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$

==> $\frac{{dv}}{{dx}} = {v^2} + 1$ ==>$\frac{{dv}}{{{v^2} + 1}} = dx$

On integrating, we get

${\tan ^{ - 1}}v = x + c$ or $v = \tan (x + c)$ ==>$x + y = \tan (x + c)$.

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