Question
Solve the following differential equation:
$+ y = \cos x - \sin x.$
$+ y = \cos x - \sin x.$
$\therefore \text{Solution is y.} e^{x} = \int(\cos x- \sin x) e^{x} dx$
$\text{Unsing} \int\text{[f(x) +f'(x)]} e^{x} + \text{c we get }$
$y.e^{x} = \cos \text{x e}^{x} + c$
$\text{or y} = \cos x + \text{c e}^{-x} $
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$\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$