Question

(dy/dx)+y=1/(1+e^(2x)) could you please solve this for me?

Answer 1

Let's first rewrite our denominator like this:\[\Large y'+y\quad=\quad \frac{1}{1+(e^{x})^2}\](It'll make sense later).
We'll introduce an integrating factor of:\[\Large \mu\quad=\quad e^{\int\limits 1\;dx}\quad=\quad e^x\]Multiplying through by our integrating factor,\[\Large e^x y'+e^x y\quad=\quad\frac{e^x}{1+(e^{x})^2}\]Left side simplifies (product rule in reverse),\[\Large (ye^x)'\quad=\quad \frac{e^x}{1+(e^{x})^2}\]

Integrating each side gives us:\[\Large ye^x\quad=\quad \int\limits \frac{e^x}{1+(e^x)^2}\;dx\]

Answer 2

To solve this integral I guess we want to apply a Trigonometric Substitution.\[\Large e^x\quad=\quad \tan \theta\]

Answer 3

\[\Large e^x\;dx\quad=\quad \sec^2\theta\;d \theta\]

Answer 4

\[\Large ye^x\quad=\quad \int \frac{\sec^2\theta}{1+(\tan \theta)^2}\;d \theta\]

Answer 5

Should simplify down really nicely from there :)
Lemme know if you're still confused.

Answer 6

You also can use
\[
u= e^x \\
du = e^x dx\\
\Large ye^x\quad=\quad \int\limits \frac{e^x}{1+(e^x)^2}\;dx=\\
\int\limits \frac{du}{1+u)^2}=\tan^{-1}(u) = \tan^{-1}\left(e^x \right) + K
\]

Answer 7

\[
y= K e^{-x} + e^{-x} \tan^{-1} \left( e^x\right)
\]

Answer 8

Oh yes, u sub is much better :) do that instead.