Solution to Separable Differential Equation/General Result

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Theorem

Consider the separable differential equation:

$\map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \dfrac {\d y} {\d x} = 0$


Its general solution is found by solving the integration:

$\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y = C$


Proof

\(\ds \map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \frac {\d y} {\d x}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x}\) \(=\) \(\ds 0\) dividing both sides by $\map {g_2} x \map {h_1} y$
\(\ds \leadsto \ \ \) \(\ds \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x\) \(=\) \(\ds \int 0 \rd x\) integrating both sides with respect to $x$
\(\ds \leadsto \ \ \) \(\ds \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x\) \(=\) \(\ds C\) Primitive of Constant
\(\ds \leadsto \ \ \) \(\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} \rd x\) \(=\) \(\ds C\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y\) \(=\) \(\ds C\) Integration by Substitution

$\blacksquare$


Sources

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