MCQ
The solution of $\frac{d y}{d x}+ y =\cos x -\sin x$ is......
- A$y e^x=\cos x+c$
- B$y e^x+e^x \cos x=c$
- ✓$y e^x=e^x \cos x+c$
- D$y^2 e^x=e^x \cos x+c$
Hint:
$\frac{d y}{d x}+ y =\cos x -\sin x$
I.F. $=e^{\int 1 d x}=e^x$
$\therefore$ the solution is $y \cdot e ^{ x }=\int(\cos x -\sin x ) e ^{ x }+ c$
$\therefore y \cdot e ^{ x }= e ^{ x } \cos x + c$
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