Differential equations

3.8 Differential equations

Writing a differential equation

A differential equation connects a quantity to its rate of change. The word "proportional" brings a constant into the equation.

Separating the variables

If the right-hand side is a function of $x$ multiplied by a function of $y$, gather each variable on its own side:

Only one constant of integration is needed, on one side.

$\int y^{-2}\,\mathrm{d}y = \int 2x\,\mathrm{d}x \quad\Rightarrow\quad -\frac{1}{y} = x^2 + c$

At $x = 0$, $y = 1$: $-1 = c$. So $-\dfrac{1}{y} = x^2 - 1$, which gives

$y = \frac{1}{1 - x^2}$

General and particular solutions

The answer with $c$ still in it is the general solution. Using an initial condition to find $c$ gives the particular solution.

Exponential growth and decay

$\int \frac{1}{N}\,\mathrm{d}N = \int k\,\mathrm{d}t \quad\Rightarrow\quad \ln N = kt + c$

Taking exponentials gives $N = \mathrm{e}^{kt + c} = A\mathrm{e}^{kt}$, where $A = \mathrm{e}^c$. At $t = 0$, $A = N_0$, so $N = N_0\mathrm{e}^{kt}$.

A problem in context

Separate and integrate:

$\int \frac{1}{\theta - 20}\,\mathrm{d}\theta = \int -k\,\mathrm{d}t \quad\Rightarrow\quad \ln(\theta - 20) = -kt + c$

At $t = 0$, $\theta = 100$, so $c = \ln 80$. Then $\theta - 20 = 80\mathrm{e}^{-kt}$.

At $t = 5$, $\theta = 60$: $40 = 80\mathrm{e}^{-5k}$, so $k = \dfrac{\ln 2}{5} = 0.139$.

$\theta = 20 + 80\mathrm{e}^{-0.139t}$

As $t$ grows, $\mathrm{e}^{-0.139t}$ approaches 0, so $\theta$ approaches 20°C: the tea cools to room temperature and never falls below it.

Combining with earlier methods

Differential equations are often used to test integration. Watch for these forms after separating:

$\int \frac{1}{y + 1}\,\mathrm{d}y = \int \frac{1}{x}\,\mathrm{d}x \quad\Rightarrow\quad \ln(y + 1) = \ln x + c$

At $x = 1$, $y = 3$: $c = \ln 4$. So $\ln(y + 1) = \ln(4x)$, and $y = 4x - 1$.