Solution to Linear First Order Ordinary Differential Equation/Proof 2
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Theorem
A linear first order ordinary differential equation in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
has the general solution:
- $\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
Proof
From the Product Rule for Derivatives:
\(\ds \frac \d {\d x} \paren {e^{\int \map P x \rd x} y}\) | \(=\) | \(\ds e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + y e^{\int \map P x \rd x} \map P x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{\int \map P x \rd x} \paren {\dfrac {\d y} {\d x} + \map P x y}\) |
Hence, multiplying $(1)$ all through by $e^{\int \map P x \rd x}$:
- $\dfrac \d {\d x} \paren {e^{\int \map P x \rd x} y} = \map Q x e^{\int \map P x \rd x}$
Integrating with respect to $x$ now gives:
- $\ds e^{\int \map P x \rd x} y = \int \map Q x e^{\int \map P x \rd x} \rd x + C$
whence we get the result by dividing by $e^{\int \map P x \rd x}$.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$