Solution of differential equation dy d x +x , ,sin^2 y = sin , y … - Vidyadip

MCQ

Solution of differential equation $\frac{dy}{d x}+x \,\,sin^2 y = sin\, y \,\,cos \,\,y$ is-

A
$tan\,\,y = (x -1) + Ce^{-x}$
✓
$cot\,\,y = (x -1) + Ce^{-x}$
C
$tan\,\,y = (x -1)e^x + C$
D
$cot\,\,y = (x -1)e^x + C$

✓

Answer

Correct option: B.

$cot\,\,y = (x -1) + Ce^{-x}$

b
$\frac{d y}{d x}+x \sin ^{2} y=\sin y \cos y$

$\cos e{c^2}y\frac{{dy}}{{dx}} + x = \cot y$

Let $-$ $cot\,y = v$

$\frac{d v}{d x}+v=x$

$\therefore \quad-$ $cot\,y \cdot {e^x} = \int x {e^x}dx$

$\Rightarrow \quad$ $cot\,y = (x - 1) + C{e^{ - x}}$

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