The solution of dy dx = (x + y) + (x + y) is - Vidyadip

MCQ

The solution of $\frac{{dy}}{{dx}} = \sin (x + y) + \cos (x + y)$ is

A
$\log \left[ {1 + \tan \left( {\frac{{x + y}}{2}} \right)} \right] + c = 0$
✓
$\log \left[ {1 + \tan \left( {\frac{{x + y}}{2}} \right)} \right] = x + c$
C
$\log \left[ {1 - \tan \left( {\frac{{x + y}}{2}} \right)} \right] = x + c$
D
None of these

✓

Answer

Correct option: B.

$\log \left[ {1 + \tan \left( {\frac{{x + y}}{2}} \right)} \right] = x + c$

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

Therefore, the differential equation reduces to

$\frac{{dv}}{{dx}} = (1 + \cos v) + \sin v$

$ = 2{\cos ^2}\frac{v}{2} + 2\sin \frac{v}{2}\cos \frac{v}{2} = 2{\cos ^2}\frac{v}{2}\left( {1 + \tan \frac{v}{2}} \right)$

==> $\int_{}^{} {\frac{{{{\sec }^2}(v/2)dv}}{{2[1 + \tan (v/2)]}}} = \int_{}^{} {dx} $

==> $\log \left[ {1 + \tan \left( {\frac{{x + y}}{2}} \right)} \right] = x + c$.

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