Video - How to Solve 1st Order Homogeneous Differential Equations (with Separation of Variables)

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:

Form of a homogeneous linear differential equation

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

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:

Form of a homogeneous linear DEQ in Leibniz notation

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

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

Homogeneous linear DEQ transformed

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

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:

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:

Calculate integral on the left side of the DEQ

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

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

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

Rearrange integrated DEQ further

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

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

Bring constant to the other side

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

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

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

Example: Solve DEQ for decay law with SoV method

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

Homogeneous first order DEQ for the decay law

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

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:

General solution of the DEQ for the decay law

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

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:

Insert initial condition into the general solution

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

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.

+ Perfect for high school and undergraduate physics students + Contains over 500 illustrated formulas on just 140 pages + Contains tables with examples and measured constants + Easy for everyone because without vectors and integrals