Linear First Order ODE/y' + y = sech x

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Theorem

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} + y = \sech x$

has the general solution:

$y = e^{-x} \paren {C + \map \ln {e^{2 x} + 1} }$


Proof

$(1)$ is a linear first order ODE with constant coefficients in the form:

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

where:

$p = 1$
$\map Q x = \sech x$


Thus:

\(\ds y\) \(=\) \(\ds e^{-x} \int e^x \sech \rd x + C e^{-x}\) Solution to Linear First Order ODE with Constant Coefficients
\(\ds \) \(=\) \(\ds e^{-x} \int \frac {2 e^x} {e^x + e^{-x} } \rd x + C e^{-x}\) Definition 1 of Hyperbolic Secant


It remains to evaluate the primitive.

Let:

\(\ds u\) \(=\) \(\ds e^x\)
\(\ds \leadsto \ \ \) \(\ds \d u\) \(=\) \(\ds e^x \rd x\)
\(\ds \leadsto \ \ \) \(\ds \) \(\) \(\ds \int \frac {2 e^x} {e^x + e^{-x} } \rd x\)
\(\ds \) \(=\) \(\ds \int \frac {2 \rd u} {u + \frac 1 u} \rd x\)
\(\ds \) \(=\) \(\ds \map \ln {u^2 + 1} + C\) Primitive of $\dfrac x {x^2 + a^2}$ for $a = 1$
\(\ds \) \(=\) \(\ds \map \ln {e^{2 x} + 1} + C\) substituting back for $u$

The result follows.

$\blacksquare$


Sources