Lesson Video: Separable Differential Equations | Nagwa Lesson Video: Separable Differential Equations | Nagwa

# Lesson Video: Separable Differential Equations Mathematics

In this video, we will learn how to identify and solve separable differential equations.

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### Video Transcript

In this video, we’ll learn the definition of a separable differential equation and how we can rewrite these to obtain equality between two integrals that we can then evaluate. To this end, it’s important you’re confident in evaluating integrals of a variety of functions — such as polynomials, trigonometric, and exponential functions — before accessing this video.

A separable equation is a first order differential equation in which the expression for d𝑦 by d𝑥 can be factored as a function of 𝑥 times a function of 𝑦. In other words, it can be written in the form d𝑦 by d𝑥 equals 𝑔 of 𝑥 times 𝑓 of 𝑦. The name separable comes from the fact that the expression on the right-hand side can be separated into a function of 𝑥 and a function of 𝑦. Equivalently, if 𝑓 of 𝑦 is not equal to zero, we can write our equation as d𝑦 by d𝑥 equals 𝑔 of 𝑥 over ℎ of 𝑦, where ℎ of 𝑦 is one over 𝑓 of 𝑦.

To solve this equation, we write it in the differential form ℎ of 𝑦 d𝑦 equals 𝑔 of 𝑥 d𝑥 Essentially, we needed to get all 𝑦s on one side of the equation and all 𝑥s on the other. And then we can integrate both sides of the equation. Now, it’s important to realize that d𝑦 by d𝑥 isn’t a fraction. But for the purposes of solving separable differential equations, we do treat it a little like one. And what we’ve done here is defined 𝑦 implicitly as a function in 𝑥. And in many cases, we’ll be able to solve for 𝑦 in terms of 𝑥. Let’s have a look at an example of how this might work.

Solve the differential equation d𝑦 by d𝑥 plus 𝑦 equals one.

A separable equation is a first order differential equation in which the expression for d𝑦 by d𝑥 can be factored as a function of 𝑥 times a function of 𝑦. In other words, it can be written in the form 𝑔 of 𝑥 times 𝑓 of 𝑦. So let’s rearrange our equation d𝑦 by d𝑥 plus 𝑦 equals one so that it’s in this form. To achieve this, we’re going to subtract 𝑦 from both sides of the equation. And we obtain d𝑦 by d𝑥 to be equal to one minus 𝑦. Now, it may not look like it, but we have achieved our aim. Our function in 𝑦 is one minus 𝑦, and our function in 𝑥 is simply one.

Then, to solve this equation, we’re going to rewrite it using differentials. Now, remember, d𝑦 by d𝑥 absolutely isn’t a fraction. But we do treat it a little like one for the purposes of this process. We begin by dividing both sides of our equation by one minus 𝑦. And we see that this is equivalent to saying one over one minus 𝑦 d𝑦 equals one d𝑥. And now, we’re ready to integrate both sides of this equation. So how do we integrate one over one minus 𝑦 with respect to 𝑦? Well, we begin by quoting the general result for the integral of one over 𝑥 with respect to 𝑥. It’s the natural log of the absolute value of 𝑥 plus some constant of integration, 𝑐.

What we’re going to do is perform a substitution for our integral. We’re going to let 𝑢 be equal to one minus 𝑦 so that d𝑢 by d𝑦 is equal to negative one. We could write this equivalently as negative d𝑢 equals d𝑦. And then we’re going to replace d𝑦 with negative d𝑢 and one minus 𝑦 with 𝑢. And we see that we now need to integrate negative one over 𝑢 with respect to 𝑢. Well, that’s negative the natural log of the absolute value of 𝑢 plus that constant of integration 𝑐. By replacing 𝑢 with one minus 𝑦, we find the integral of one over one minus 𝑦 to be the negative natural log of the absolute value of one minus 𝑦 plus a constant of integration which I’m going to call 𝑎.

When we integrate one with respect to 𝑥, it’s a little more straightforward. We get 𝑥 plus a second constant of integration. Let’s call that 𝑏. Let’s subtract 𝑎 from both sides of our equation and multiply through by negative one. That gives us the natural log of the absolute value of one minus 𝑦 equals negative 𝑥 plus 𝑐 one. 𝑐 one is a new constant, and it’s achieved by subtracting 𝑎 from 𝑏 and multiplying by negative one. And then, we notice that we can raise both sides of this equation as a power of 𝑒. So we obtain the absolute value of one minus 𝑦 to be equal to 𝑒 to the power of negative 𝑥 plus 𝑐 one.

By using the laws of exponents, though, we see that we can rewrite 𝑒 to the power of negative 𝑥 plus 𝑐 one as 𝑒 to the power of negative 𝑥 times 𝑒 to the power of 𝑐 one. But, of course, 𝑒 to the power of 𝑐 one is itself a constant. Let’s call that 𝑐 two. And we see we can rewrite the right-hand side as 𝑐 two times 𝑒 to the power of negative 𝑥. Well, of course, we’re dealing with the absolute value of one minus 𝑦. And we can say that this means that one minus 𝑦 could be equal to positive 𝑐 two times 𝑒 to the power of negative 𝑥 or negative 𝑐 two times 𝑒 to the power of negative 𝑥. But 𝑐 two is a constant, so we don’t actually need to write that.

In our final step, we’re going to add 𝑦 to both sides of the equation and then subtract 𝑐 two 𝑒 to the power of negative 𝑥. That gives us 𝑦 equals one minus 𝑐 two times 𝑒 to the power of negative 𝑥. But of course, once again, since 𝑐 two is a constant, we can change this to plus 𝑐. And we see that 𝑦 equals one plus 𝑐 times 𝑒 to the power of negative 𝑥 is the solution to our separable differential equation.

We’ll now have a look at an example which requires some further techniques for integration.

Find a relation between 𝑦 and 𝑥 given that 𝑥𝑦 times 𝑦 prime is equal to 𝑥 squared minus five.

Now, this first step isn’t entirely necessary. But it can make it a little easier to see what to do next. We write 𝑦 prime using Leibniz notation. And we see that 𝑥𝑦 d𝑦 by d𝑥 is equal to 𝑥 squared minus five. And then we recall that a separable differential equation is one in which the expression for d𝑦 by d𝑥 can be written as some function of 𝑥 times some function of 𝑦. Now, in fact, if we divide both sides of our equation by 𝑥𝑦, we see that we can achieve this. We get d𝑦 by d𝑥 equals 𝑥 squared minus five over 𝑥𝑦 or 𝑥 squared minus five over 𝑥 times one over 𝑦. And that’s great because 𝑔 of 𝑥 is, therefore, 𝑥 squared minus five over 𝑥. And our function of 𝑦 is one over 𝑦.

Now, in fact, we just performed this to demonstrate that we did indeed have a separable differential equation. We could’ve kept 𝑦 on the left-hand side as shown. And then, we perform this rather strange step. d𝑦 by d𝑥 isn’t a fraction, but we do treat it a little like one. And we say that 𝑦 d𝑦 equals 𝑥 squared minus five over 𝑥 d𝑥. Our next step is to integrate both sides of this equation. Remember, to integrate a polynomial term whose exponent is not equal to negative one, we add one to the exponent and divide by that new number. So the integral of 𝑦 is 𝑦 squared over two plus some constant of integration 𝑎.

Then, it might look like we need to perform some sort of substitution to evaluate this integral. But actually, if we separate our fraction into 𝑥 squared over 𝑥 minus five over 𝑥, we find that we need to integrate 𝑥 minus five over 𝑥 with respect to 𝑥. Then, we recall the general result for the integral of one over 𝑥. It’s the natural log of the absolute value of 𝑥 plus 𝑐. And so, when we integrate the right-hand side of our equation, we get 𝑥 squared over two minus five times the natural log of the absolute value of 𝑥 plus some second constant of integration which I’ve called 𝑏.

Our last step is to subtract 𝑎 from 𝑏 and then multiply the entire equation by two. When we do, we find that 𝑦 squared equals 𝑥 squared minus 10 times the natural log of the absolute value of 𝑥 plus 𝑐. This 𝑐 is a different constant achieved by subtracting 𝑎 from 𝑏 and then multiplying by two. And so, given our differential equation, we’ve found a relation between 𝑦 and 𝑥. 𝑦 squared equals 𝑥 squared minus 10 times the natural log of the absolute value of 𝑥 plus 𝑐.

In this next example, we’ll see how some clever factoring can create a much simpler expression to separate.

Find a relation between 𝑢 and 𝑡 given that d𝑢 by d𝑡 equals one plus 𝑡 to the fourth power over 𝑢𝑡 squared plus 𝑢 to the fourth power times 𝑡 squared.

Now, this differential equation might look really nasty. But it is, in fact, a separable differential equation. In this case, that’s one where an expression for d𝑢 by d𝑡 can be written as some function of 𝑢 times some function of 𝑡. So how are we going to achieve that? Well, we begin by factoring the denominator by 𝑡 squared. And we find that d𝑢 by d𝑡 is equal to one plus 𝑡 to the fourth power over 𝑡 squared times 𝑢 plus 𝑢 to the fourth power. We can now write this as some function of 𝑡 times some function of 𝑢. It’s one plus 𝑡 to the fourth power over 𝑡 squared times one over 𝑢 plus 𝑢 to the fourth power.

Let’s begin by multiplying both sides of this equation by 𝑢 plus 𝑢 to the fourth power. Then, we recall that whilst d𝑢 by d𝑡 is not a fraction, we treat it a little like one. And we can say that 𝑢 plus 𝑢 to the fourth power d𝑢 is equal to one plus 𝑡 to the fourth power over 𝑡 squared d𝑡. And that’s great because we’re now ready to integrate both sides of our equation. The left-hand side is quite straightforward to integrate. The integral of 𝑢 is 𝑢 squared over two. And the integral of 𝑢 to the fourth power is 𝑢 to the fifth power over five. Don’t forget, since this is an indefinite integral, we need that constant of integration 𝑎.

On the right-hand side, we’re going to rewrite our integrand as one over 𝑡 squared plus 𝑡 to the fourth power over 𝑡 squared. And that simplifies to one over 𝑡 squared plus 𝑡 squared. But, of course, one over 𝑡 squared is the same as 𝑡 to the power of negative two. And we can now integrate as normal. When we integrate 𝑡 to the power of negative two, we add one to the exponent to get negative one and then we divide by that number. 𝑡 squared becomes 𝑡 cubed over three. And we have a second constant of integration; I’ve called that 𝑏. Now, of course, we can rewrite 𝑡 to the power of negative one over negative one as negative one over 𝑡.

Our final step is to subtract our constant 𝑎 from both sides of the equation. That gives us a new constant 𝑐. That’s just the difference between 𝑏 and 𝑎. So we’ve found a relation between 𝑢 and 𝑡. It’s 𝑢 to the fifth power over five plus 𝑢 squared over two equals negative one over 𝑡 plus two cubed over three plus 𝑐.

In our final example, we’ll see how this process works for exponential functions.

Solve the differential equation d𝑧 by d𝑡 plus 𝑒 to the power of two 𝑡 plus two 𝑧 equals zero.

Remember, a separable differential equation is one for which the expression for d𝑧 by d𝑡 can be expressed as some function of 𝑧 times some function of 𝑡. So how exactly are we going to achieve that for our equation? Well, we’re going to begin by subtracting 𝑒 the power of two 𝑡 plus two 𝑧 from both sides of our equation. We then recall that the laws of exponents tell us that 𝑥 to the power of 𝑎 plus 𝑏 can be written as 𝑥 to the power of 𝑎 times 𝑥 to the power of 𝑏. So we can write negative 𝑒 to the power of two 𝑡 plus two 𝑧 as negative 𝑒 to the power of two 𝑡 times 𝑒 to the power of two 𝑧.

Now, of course, d𝑧 by d𝑡 isn’t a fraction, but we treat it a little like one. And we can say that this is equivalent to one over 𝑒 to the power of two 𝑧 d𝑧 equals negative 𝑒 to the two 𝑡 d𝑡. And this is great because we’re now ready to integrate both sides. It can be simpler to express one over 𝑒 to the power of two 𝑧 as 𝑒 to the power of negative two 𝑧. And then, we quote the general result for the integral of 𝑒 to the power of 𝑘𝑥 for some constant 𝑘. It’s 𝑒 to the power of 𝑘𝑥 over 𝑘 plus 𝑐. So this means the integral of 𝑒 to the power of negative two 𝑧 is negative 𝑒 to the power of negative two 𝑧 over two. And the integral of negative 𝑒 to the two 𝑡 is negative 𝑒 to the two 𝑡 over two.

Our next step is to subtract 𝑐 one from both sides of the equation and then multiply through by negative two. Remember, in solving our differential equation, ideally, we want an equation for 𝑧 in terms of 𝑡, So we find that 𝑒 to the power of negative two 𝑧 is equal to 𝑒 to the power of two 𝑡 plus 𝑐 three. 𝑐 three is a new constant obtained by subtracting 𝑐 one from 𝑐 two and then multiplying by negative two. To solve for 𝑧, we find the natural log of both sides of this equation. But the natural log of 𝑒 to the power of negative two 𝑧 is just negative two 𝑧. So our final step is to divide through by negative two. And we’ve solved our differential equation. 𝑧 is equal to negative one-half times the natural log of 𝑒 to the power of two 𝑡 plus some constant; let’s call that 𝑐.

In this video, we’ve seen that a separable equation is a first order differential equation of the form d𝑦 by d𝑥 equals some function of 𝑥 times some function of 𝑦. We saw that to solve these kinds of equations, we separate all our 𝑥s onto one side and all our 𝑦s onto the other, and then we integrate. Finally, we emphasized that whilst we do treat d𝑦 by d𝑥 a little like a fraction in this method, d𝑦 by d𝑥 is absolutely not a fraction.