Solve the following differential equation:
\frac{dy}{dx}+y= \cos x-\sin x

 

 

 

 
 
 
 
 

Answers (1)

The equation is of the form \frac{dy}{dx}+p\left ( x \right )y= Q\left ( x \right )
where, p\left ( x \right )= 1,Q\left ( x \right )= \cos x-\sin x
\therefore IF={e}^{ \int 1dx}= e^{x}
The solution is given by e^{x}y= \int e^{x}\left ( \cos x-\sin x \right )dx+c
e^{x}y= e^{x}\left ( \cos x \right )+c \:\: or
y= \cos x+ce^{-x}\left [ \because \int e^{x}\left \{ f\left ( x \right )+f'\left ( x \right )dx=e^{x}f\left ( x \right )+c \right \} \right ]

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