Separation of Variables (SoV) and How to Solve Homogeneous DEQ's of 1st Order

by Alexander FufaeV

Separation of variables is suitable for ordinary 1st order DEQ's which are homogeneous.

The method of separation of variables is suited for:

ordinary DEQ of 1st order,
which is linear
and homogeneous.

Remember that if a DEQ is homogeneous, it is also linear. This type of DEQ has the form:

$$ \begin{align} y' ~+~ K(x)\,y ~=~ 0 \end{align} $$

Here the coefficient $K$ must not necessarily constant, but can also depend on $x$! Also note that before the first derivative $y'$ the coefficient must be equal to 1. If this is not the case for you, then you simply have to divide the whole equation by the coefficient which is in front of $y'$. Then you have the right form.

In this solving method, $y$ and $x$ are considered as two variables and separated from each other by bringing $y$ to one side and $x$ to the other side of the equation. The Leibniz notation of the differential equation is best suited for this purpose:

$$ \begin{align} \frac{\text{d}y}{\text{d}x} ~+~ K(x)\,y ~=~ 0 \end{align} $$

Bring $K(x)\,y$ to the right hand side:

$$ \begin{align} \frac{\text{d}y}{\text{d}x} ~=~ -K(x)\,y \end{align} $$

Multiply the equation by $ \text{d}x $ and then divide the equation by $y$. This way you have only $y$ dependence on the left side and only $x$ dependence on the right side:

$$ \begin{align} \frac{1}{y} \, \text{d}y ~=~ -K(x) \, \text{d}x \end{align} $$

Now you can integrate over $y$ on the left hand side and over $x$ on the right hand side:

$$ \begin{align} \int \frac{1}{y} \, \text{d}y ~=~ -\int K(x) \, \text{d}x \end{align} $$

Integrating $ 1 / y $ gives the natural logarithm of $y$. It's best to know this by heart, because you will often encounter such an integral. Also, don't forget the integration constant! Let's call it $A$ for example:

$$ \begin{align} \ln(y) ~+~ A ~=~ -\int K(x) \, \text{d}x \end{align} $$

Now you have to solve for the function $y$. Use the exponential function $\mathrm{e}^{...}$ on both sides:

$$ \begin{align} \mathrm{e}^{\ln(y) ~+~ A} ~=~ \mathrm{e}^{-\int K(x) \, \text{d}x} \end{align} $$

You can split the sum in the exponential term on the left into a product, where $\mathrm{e}^{\ln(y)}$ is simply $y$:

$$ \begin{align} y \, \mathrm{e}^{A} ~=~ \mathrm{e}^{-\int K(x) \, \text{d}x} \end{align} $$

Bring the constant $\mathrm{e}^{A}$ to the right side:

$$ \begin{align} y ~=~ \frac{1}{\mathrm{e}^{A}} \, \mathrm{e}^{-\int K(x) \, \text{d}x} \end{align} $$

Rename $ \frac{1}{\mathrm{e}^{A}} $ to a new constant $C$. As a result, you get a general solution formula that you can always use to solve homogeneous linear differential equations. You don't have to apply the separation of variables method again and again, but you can use the solution formula directly:

Solution formula for ordinary homogeneous DEQ of 1st order

$$ \begin{align} y ~=~ C\, \mathrm{e}^{-\int K(x) \, \text{d}x} \end{align} $$

Example: Solve DEQ for decay law with SoV method

Let's look at the differential equation for the radioactive decay law:

$$ \begin{align} \frac{\text{d}N}{\text{d}t} ~+~ \lambda \, N ~=~ 0 \end{align} $$

In this case the searched function $y$ is the number of not yet decayed atomic nuclei $N$ and the variable $x$ corresponds to the time $t$. And the coefficient $K$ in this case is a decay constant $\lambda$. They are just different letters. The type of the differential equation is the same! According to the solution formula you have to integrate the coefficient, i.e. the decay constant over $t$. Integrating a constant just gives $t$. And you have the general solution for the decay law:

$$ \begin{align} N ~=~ C\, \mathrm{e}^{-\lambda \, t} \end{align} $$

Exponential decay of the number of atomic nuclei in the decay law.

Now you know the qualitative behavior of the physical process, namely that atomic nuclei decay exponentially. But you cannot say concretely yet, how many nuclei have already decayed after a certain period of time. This is because you don't know the constant $C$ yet. After all, in the decay law, $C$ gives the number of atomic nuclei that were present at the beginning of your observation. So you need an initial condition as additional information to the differential equation. It could be for example like this: $ N(0) = 1000 $. That means, at the time $t = 0 $ there were 1000 atomic nuclei. Inserting this initial condition results in this equation:

$$ \begin{align} N(0) ~=~ 1000 ~=~ C\, \mathrm{e}^{-\lambda \cdot 0} \end{align} $$

So $ C $ must be 1000:

$$ \begin{align} N ~=~ 1000\, \mathrm{e}^{-\lambda \, t} \end{align} $$

Now you can insert an arbitrary point of time $t$ and find out how many not decayed atomic nuclei $N(t) $ are still there.

Now you know how to solve simple homogeneous linear differential equations of 1st order. In the next lesson, we will look at how to crack inhomogeneous DEQ's with the "variation of constants" method.