Linear First Order ODE/y' + y = sech x
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Theorem
- $(1): \quad \dfrac {\d y} {\d x} + y = \sech x$
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
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $4$