Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)
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Theorem
- $(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$
has the general solution:
- $y = e^{-x} \map \arctan {e^x} + C e^{-x}$
Proof
$(1)$ 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:
\(\ds e^x y\) | \(=\) | \(\ds \int \frac {1 \rd u} {1 + u^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \arctan u + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \arctan {e^x} + C\) |
or:
- $y = e^{-x} \map \arctan {e^x} + C e^{-x}$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $2 \ \text{(b)}$