Question
Show that the given equation is a solution of the given differential equation.$$y^{\prime}+y=2 \cos x, \quad y=\sin x+\cos x-e^{-x}$$
Step 1
The derivative of $y=\sin x+\cos x-e^{-x}$ is given by: $$y'=\cos x-\sin x+e^{-x}$$ Show more…
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