Separable ODE — Physica

Equivalent forms

\frac{dy}{y} = -k\,dx \Rightarrow y = y_0 e^{-kx}

\int \frac{dy}{g(y)} = \int f(x)\,dx + C

The equation whose solution — the exponential — shows up wherever a quantity's rate of change is proportional to the quantity itself.

Symbol	Name	SI unit	Dimension	Range
$k$	Rate constant Decay/growth rate in dy/dx = -k y	1	1	-1.5 – 1.5
$y_0$	Initial value Starting height of the solution	1	1	0.3 – 3
$x$	Independent variable Position/time	1	1	0 – 6
$y$	Solutionoutput y(x) = y_0 e^{-kx}	1	1	0 – 3

Unit systems

Symbol	SI	CGS	Imperial
$x$	1	1	1

Where it holds

Applies whenever the RHS factorizes as f(x)g(y). Care needed where

g(y)=0

(equilibrium/singular solutions) since dividing by g is illegal there.

Dimensional analysis

dy/dx ([y]/[x])

f(x)g(y)

For

dy/dx=-ky

, k carries 1/[x] so ky has [y]/[x], matching the derivative; the exponent kx is dimensionless.

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Discovery

Leibniz & Bernoulli · 1691

Leibniz introduced separation of variables around 1691 and Johann Bernoulli developed it for early differential equations. It is the first solving technique every student meets and the workhorse behind exponential growth/decay models across physics, chemistry, and biology.

Try this

Radioactive decay, cooling coffee, viral spread, charging capacitors — a huge slice of change in nature obeys dy/dx = f(x)g(y). Split the variables to opposite sides and the whole thing integrates in one move.

Follow a solution curve threading through a slope field: at every point its tangent matches the little arrow, and changing the starting height slides you onto a different member of the family.

Research status: stable

Real-world applications

Radioactive decay and carbon dating
Newton's law of cooling
RC/RL circuit charging and discharging
Logistic and unlimited population growth

Common misconceptions

You can always separate — only if the RHS factorizes into f(x)g(y).
dy/dx is a fraction you can split — it's shorthand; the manipulation is justified by the chain rule, not literal division.
The constant C is optional — it encodes the initial condition and the whole solution family.

Experimental verification

Radioactive-decay counting (half-life curves), Newton's-law-of-cooling thermometer traces, and RC-circuit discharge voltages all match y_0 e^{-kx} to high precision — textbook confirmations of the separable solution.

Derivation

From

dy/dx = f(x)g(y)

, divide by g(y) (where nonzero) and multiply by dx:

dy/g(y) = f(x)dx

.

Integrate both sides independently:

\int dy/g(y) = \int f(x)dx + C

.

For

dy/dx=-ky

this gives

\ln y = -kx + C

, i.e.

y = y_0\,\mathrm{e}^{-kx}

.

Limiting cases

k > 0⟶ exponential decayy shrinks toward zero — radioactive decay, cooling, discharge.

k < 0⟶ exponential growthy explodes — unchecked populations, chain reactions.

k = 0

⟶ constantNo driving rate; y stays at y_0.

What if…

What if

g(y)=0

somewhere?

That constant y is an equilibrium (singular) solution the division step hides — add it back explicitly.

What if the RHS won't factor?

Try an integrating factor (linear ODE) or substitution; separation only works for factorable right-hand sides.

1

Solve dy/dx = -k y

Given ·

y(0):: y_0

Find · y(x)

Steps

$dy/y = -k dx$
$\ln y = -kx + C$
$y = y_0\,\mathrm{e}^{-kx}$ with C from y(0)

Result ·

y = y_0\,\mathrm{e}^{-kx}

2

Half-life of decay

Given ·

y:: y_0/2

Find · x at half

Steps

$\tfrac{1}{2} = e^{-kx}$
$\ln \tfrac{1}{2} = -kx$
$x = \ln 2/k$

Result ·

x = \ln 2 / k

first order linear odeintegrating factorFundamental Theorem of Calculus→exact ode