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