Linear First Order ODE/y' + 2y = cos x
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Theorem
- $(1): \quad \dfrac {\d y} {\d x} + 2 y = \cos x$
has the general solution:
- $y = \dfrac {2 \cos x + \sin x} 5 + C e^{-2 x}$
Proof
$(1)$ is a linear first order ODE in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
- $\map P x = 2$
- $\map Q x = \cos x$
Thus:
\(\ds \int \map P x \rd x\) | \(=\) | \(\ds \int 2 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int P \rd x}\) | \(=\) | \(\ds e^{2 x}\) |
Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
\(\ds \map {\dfrac {\d} {\d x} } {y e^{2 x} }\) | \(=\) | \(\ds e^{2 x} \cos x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y e^{2 x}\) | \(=\) | \(\ds \int e^{2 x} \cos x \rd x\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{2 x} \paren {2 \cos x + \sin x} } {2^2 + 1^2} + C\) | Primitive of Exponential of $e^{a x} \cos b x$ for $a = 2, b = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{2 x} \paren {2 \cos x + \sin x} } 5 + C\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \dfrac {2 \cos x + \sin x} 5 + C e^{-2 x}\) |
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $3$