Separation of variables

This technique can be used to solve more complicated equations, such as those where the rate of change is a product of functions of the dependent and of the independent variables, i.e.

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(x)}{g'(y)}$$

To solve such an equation, we first ``separate the variables'' as shown.

$$g'(y) \mathrm{d}y = f'(x) \mathrm{d}x$$

We now integrate both sides to obtain the solution

$$\begin{eqnarray*} \int g'(y) \mathrm{d}y & = & \int f'(x) \mathrm{d}x \\ g(y) & = & f(x) + c \\ y & = & g^{-1} (f(x) + c) \end{eqnarray*}$$

We then proceed to find the particular solution, as before.

As an example, we will use the equation, arising in physics, modelling radioactive decay. Here, is a constant known in physics as the ``decay constant''. represents the number of undecayed nuclei, and is the time in seconds. The equation is

$$\frac{\mathrm{d}N}{\mathrm{d}t} = - \lambda N$$

(The negative sign shows that is decreasing.) To solve the equation, we separate the variables to obtain

$$\frac{\mathrm{d}N}{N} = - \lambda \mathrm{d}t$$

We now integrate both sides to find the general solution.

$$\begin{eqnarray*} \int \frac{\mathrm{d}N}{N} & = & - \lambda \int \mathrm{d}t \\... ... \mathrm{e}^{-\lambda t} \\ N & = & N_0 \mathrm{e}^{-\lambda t} \end{eqnarray*}$$

Here, represents the initial number of undecayed nuclei - the number at .