# 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

\(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds -k \paren {y - y_a}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d y} {y - y_a}\) | \(=\) | \(\ds -\int k \rd x\) | Separation of Variables | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map \ln {y - y_a}\) | \(=\) | \(\ds -k x + C_1\) | 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:

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

$\blacksquare$