Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)

Theorem

$(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$

$y = e^{-x} \map \arctan {e^x} + C e^{-x}$

$(1)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where $\map P x = 1$.

Thus:

$\ds \int \map P x \rd x$

$=$

$\ds \int \rd x$

$\ds $

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds e^{\int P \rd x}$

$=$

$\ds e^x$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac \d {\d x} } {e^x y} = \dfrac {e^x} {1 + e^{2 x} }$

and the general solution becomes:

$\ds y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$

The integral on the right hand side can be solved by substituting:

$u = e^x \implies \d u = e^x \rd x$

and so:

$\ds e^x y$

$=$

$\ds \int \frac {1 \rd u} {1 + u^2}$

$\ds $

$=$

$\ds \map \arctan u + C$

$\ds $

$=$

$\ds \map \arctan {e^x} + C$

or:

$y = e^{-x} \map \arctan {e^x} + C e^{-x}$

$\blacksquare$