Decay Equation

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Theorem

The first order ordinary differential equation:

$\dfrac {\d y} {\d x} = k \paren {y_a - y}$

where $k \in \R: k > 0$

has the general solution:

$y = y_a + C e^{-k x}$

where $C$ is an arbitrary constant.

If $y = y_0$ at $x = 0$, then:

$y = y_a + \paren {y_0 - y_a} e^{-k x}$

This differential equation is known as the decay equation.

Proof

\(\ds \frac {\d y} {\d x}\)

\(=\)

\(\ds -k \paren {y - y_a}\)

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\d y} {y - y_a}\)

\(=\)

\(\ds -\int k \rd x\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds \map \ln {y - y_a}\)

\(=\)

\(\ds -k x + C_1\)

Primitive of Reciprocal and Derivatives of Function of $a x + b$

\(\ds \leadsto \ \ \)

\(\ds y - y_a\)

\(=\)

\(\ds e^{-k x + C_1}\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds y_a + C e^{-k x}\)

where we put $C = e^{C_1}$

This is our general solution.

$\Box$

Suppose we have the initial condition:

$y = y_0$ when $x = 0$

Then:

$y_0 = y_a + C e^{-k \cdot 0} = y_a + C$

and so:

$C = y_0 - y_a$

Hence the solution:

$y = y_a + \paren {y_0 - y_a} e^{-k x}$

$\blacksquare$

Also see

• Newton's Law of Cooling
• First-Order Reaction