**Separable differential equations**

In some ways these are the easiest differential equations to solve in theory, though in practice the final step (that of integrating) may be difficult or impossible. A separable differential equation is one of the form:

where and are any functions of and respectively. For instance:

is of this form where and . The reason that these equations are simple in theory is because we can rearrange them to be:

ie. we have all the stuff on one side and all the stuff on the other and then we can integrate both sides:

and that’s it. As long as you can do the integrals, you can get a function in terms of . let’s look at some examples:

gives the following integral:

and so:

here we have one constant of integration from each side of the interval, but because they are just constants, we can put them into one constant and call it :

we can then rearrange this to give:

We can then call just a constant, and let’s call it because we can see that when is zero, is just going to be given by this constant:

This has two solutions (one where is positive, and one where it is negative), so we can choose one of them, depending on the initial condition for . If we are told that for then:

If we are told that for , then:

so we have found the general set of solutions (and here, two particular solutions) to the initial differential equation.

Given that you can already integrate, basically that is all you need to know about solving separable differential equations, but let’s take another couple of examples:

given the initial condition that at , , ie. . We rearrange the equation to get the integral:

This gives (taking the two integration constants and writing them as one:

so , where we have absorbed a factor of 2 into . Now we want to find the value of given our initial condition, so we set and :

so and so the particular solution is:

In the figure below we have plotted this solution, along with a number of solutions with other initial conditions.

figure:

Lets look at another example:

with initial condition . Again, we rearrange the equation to be an integral:

and so giving the solution:

we plug in and to get:

which is solved by . So the particular solution that we are interested in is:

Note that this only makes sense for which has two solutions: and . Only the first of these contains the point and so this is the branch that we are interested in, so the solution is given above and is valid for . The particular solution along with several other solutions are shown here:

How does the integration tie in with Euler’s method. Well, in Euler’s method we are slowly adding up more and more change as we move along the direction and it is telling us the total that we have got to at any one value of . Integration is just adding up and so Euler’s method is just the adding up of the little bits in discrete steps rather than as a continual function as in an interval.

**Reminder: A way of thinking about these equation**

Differential equations seem a bit mysterious at first, but we really can take them as extensions of algebraic equations. When you are given the equation:

You are being asked to find that number (or those numbers) whose square is equal to 4 plus itself. Similarly when you are given then differential equation:

You are being asked to find the family of functions whose gradients are equal to their value plus the square of the value at any one point. You should be able to go anywhere on one of these curves and the relationship between , and the gradient of the line at that point should satisfy the equation. How do you know that it does? Well, just plug in the function you believe to be a solution into the differential equation and if it does indeed give you zero then you’ve found a solution.

Valentine ChisangoSeptember 13, 2015 at 11:09 pmIn the sentence before the first figure you have “In figure the figure below”

Jonathan ShockSeptember 14, 2015 at 12:00 amCheers – good spot.

J

Abigail MallowsSeptember 15, 2015 at 9:16 pmHello Dr Shock

There is an integral sign missing in the Right hand side of the integration of the example dy/dx=e^(-y)*(2x-4).

Jonathan ShockSeptember 16, 2015 at 5:47 amCheers Abigail!

Mathemafrica - UCT MAM1000 lecture notes subject linksOctober 4, 2015 at 10:36 am[…] MAM1000 lecture notes part 35 – Differential equations – separable differential equation… […]

DimitriosAugust 16, 2022 at 8:56 amHi Dr Shock,

I think the sentence beneath the last graph/figure reads:

“Solutions to the differential equation” \frac{dy}{dt}=e^{-y}(2x-4)

I think “dt” should be “dx”

Jonathan ShockAugust 16, 2022 at 11:39 amWell spotted, thank you!