Separation of Variables/General Result

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Theorem

Suppose a first order ordinary differential equation can be expressible in this form:

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

Then the equation is said to have separable variables, or be separable.


Its general solution is found by solving the integration:

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


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 Integrals
\(\ds \leadsto \ \ \) \(\ds \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y\) \(=\) \(\ds C\) Integration by Substitution

$\blacksquare$


Sources