Solution to Linear First Order Ordinary Differential Equation/Proof 2

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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:

$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$


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 $e^{\int \map P x \rd x}$.