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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.

\begin{displaymath}
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(x)}{g'(y)}
\end{displaymath}

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

\begin{displaymath}
g'(y) \mathrm{d}y = f'(x) \mathrm{d}x
\end{displaymath}

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, \( \lambda \) is a constant known in physics as the ``decay constant''. \( N \) represents the number of undecayed nuclei, and \( t \) is the time in seconds. The equation is

\begin{displaymath}
\frac{\mathrm{d}N}{\mathrm{d}t} = - \lambda N
\end{displaymath}

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

\begin{displaymath}
\frac{\mathrm{d}N}{N} = - \lambda \mathrm{d}t
\end{displaymath}

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, \( N_0 \) represents the initial number of undecayed nuclei - the number at \( t = 0 \).
next up previous
Next: Copyright Up: Differential equations Previous: Direct integration
Alexander Frolkin 2001-03-13