Solution to Separable Differential Equation

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Definition

Consider the separable differential equation:

$\dfrac {\d y} {\d x} = \map g x \map h y$

Its general solution is found by solving the integration:

$\ds \int \frac {\d y} {\map h y} = \int \map g x \rd x + C$


This technique is generally known as Separation of Variables.


General Result

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

Dividing both sides by $\map h y$, we get:

$\dfrac 1 {\map h y} \dfrac {\d y} {\d x} = \map g x$

Integrating both sides with respect to $x$, we get:

$\ds \int \frac 1 {\map h y} \frac {\d y} {\d x} \rd x = \int \map g x \rd x$

which, from Integration by Substitution, reduces to the result.

The arbitrary constant $C$ appears during the integration process.

$\blacksquare$


Mnemonic Device


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As derivatives are not fractions, the following is a mnemonic device only.

This is an an abuse of notation that is likely to make some Calculus professors upset.

But it's useful.

\(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \map g x \map h y\)
\(\ds \leadsto \ \ \) \(\ds \d y\) \(=\) \(\ds \map g x \map h y \rd x\) solve for $\d y$
\(\ds \leadsto \ \ \) \(\ds \frac 1 {\map h y} \rd y\) \(=\) \(\ds \map g x \rd x\) collecting like terms on each side


Examples

Arbitrary Example $1$

Consider the first order ODE:

$(1): \quad \map {\dfrac \d {\d x} } {\map f x} = 3 x$

where we are given that $\map f 1 = 2$.

The particular solution to $(1)$ is:

$\map f x = \dfrac {3 x^2 + 1} 2$


Arbitrary Example $2$

Solution to Separable Differential Equation/Examples/Arbitrary Example 2

Sources

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