Solveeit Logo
Login

Question

Question: The solution of the differential equation \(( \sin x + \cos x ) d y + ( \cos x - \sin x ) d x = 0\)...

The solution of the differential equation

(sin⁡x+cos⁡x)dy+(cos⁡x−sin⁡x)dx=0( \sin x + \cos x ) d y + ( \cos x - \sin x ) d x = 0is

A

ex(sin⁡x+cos⁡x)+c=0e ^ { x } ( \sin x + \cos x ) + c = 0

B

ey(sin⁡x+cos⁡x)=ce ^ { y } ( \sin x + \cos x ) = c

C

ey(cos⁡x−sin⁡x)=ce ^ { y } ( \cos x - \sin x ) = c

D

ex(sin⁡x−cos⁡x)=ce ^ { x } ( \sin x - \cos x ) = c

Answer

ey(sin⁡x+cos⁡x)=ce ^ { y } ( \sin x + \cos x ) = c

Explanation

Solution

dydx=−cos⁡x−sin⁡xsin⁡x+cos⁡x\frac { d y } { d x } = - \frac { \cos x - \sin x } { \sin x + \cos x } ⇒ dy=−(cos⁡x−sin⁡xsin⁡x+cos⁡x)dxd y = - \left( \frac { \cos x - \sin x } { \sin x + \cos x } \right) d x

On integrating both sides, we get

⇒ y=−log⁡(sin⁡x+cos⁡x)+log⁡cy = - \log ( \sin x + \cos x ) + \log c

⇒ y=log⁡(csin⁡x+cos⁡x)y = \log \left( \frac { c } { \sin x + \cos x } \right) ⇒ ey(sin⁡x+cos⁡x)=ce ^ { y } ( \sin x + \cos x ) = c.