# Separation of Variables/General Result

## 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

$\blacksquare$