Question
In Exercises $5-8,$ show that each function is a solution of the given initial value problem.$$y^{\prime}+y=\frac{2}{1+4 e^{2 x}} \quad y(-\ln 2)=\frac{\pi}{2} \quad y=e^{-x} \tan ^{-1}\left(2 e^{x}\right)$$
Step 1
Using the chain rule, we get $$y' = -e^{-x} \tan ^{-1}\left(2 e^{x}\right) + \frac{2e^{-x}}{1+(2e^{x})^2}.$$ Show more…
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