Let y=y(x) solve (1+e^2x)dydx+y=1, passing through (0,2).

Question

Let $y=y(x)$ solve $(1+e^{2x})\dfrac{dy}{dx}+y=1$, passing through $\left(0,\dfrac{\pi}{2}ight)$. Then $\lim_{x	o\infty}e^xy(x)$ is:

Accepted Answer

$$\frac{dy}{dx}+y = \frac{1}{1+e^{2x}}.$$

I.F.$=e^x$. $e^xy = \tan^{-1}(e^x)+C$. Using $\left(0,\dfrac{\pi}{2}\right)$: $\dfrac{\pi}{2}=\dfrac{\pi}{4}+C \Rightarrow C=\dfrac{\pi}{4}$.

$$\lim_{x\to\infty}e^xy(x) = \lim_{x\to\infty}\left[\tan^{-1}(e^x)+\frac{\pi}{4}\right] = \frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}.$$

Solutions · 1