Video: Initial Value Problems

In this video, we will learn how to find a specific solution to a separable differential equation given an initial value.

17:20

Video Transcript

Initial Value Problems

In this video, we’ll practise finding a specific solution to a separable differential equation when given an initial value.

To begin, we remind ourselves of the definition of a separable differential equation. This is any differential equation that can be written in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. Looking at the right-hand side of this equation, if we cannot express it as a function of only π‘₯ multiplied by a function of only 𝑦, then it is not classed as a separable differential equation. To find solutions to this type of equation, we can use the following method.

Here, we divide both sides by the function 𝑓 of 𝑦. To make our notation easier to follow, we can define a new function β„Ž of 𝑦, which is equal to one over 𝑓 of 𝑦. Also, a quick side note, when 𝑓 of 𝑦 equals zero, we’ll have to divide by zero. And hence, the solution will be lost. But we won’t be worrying too much about this during the video.

Okay, our next step will be to integrate both sides of our equation with respect to π‘₯. Now, the left-hand side of the equation might look a little strange to us. But one thing we might recognize is that the chain rule can be applied to give us that d𝑦 by dπ‘₯, dπ‘₯ is equal to d𝑦. We can use this to give us the equation that the integral of β„Ž of 𝑦 with respect to 𝑦 is equal to the integral of 𝑔 of π‘₯ with respect to π‘₯. As an aside, let us rewind a little bit. You will often see this method represented in the following way. At the stage where we had β„Ž of 𝑦 d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯, we would treat d𝑦 by dπ‘₯ in a similar fashion to a fraction, allowing us to give the equivalent statement β„Ž of 𝑦 d𝑦 is equal to 𝑔 of π‘₯ dπ‘₯.

We would now integrate both sides. And hopefully, you should be able to see that we have arrived at the same stage. You should always be aware that these steps shown on the right should be treated as a shorthand for the method that we first showed. You often see this because it uses fewer lines of working. But always be aware of the more rigorous steps that are actually going on in the background. For example, in our shorthand, we have ambiguously integrated both sides of our equation. Whereas more rigorously, we are integrating both sides with respect to π‘₯. And the result just so happens to be an integral on the left-hand side with respect to 𝑦.

Okay, let’s tidy things up and continue with our method. Integrating β„Ž of 𝑦 with respect to 𝑦 and 𝑔 of π‘₯ with respect to π‘₯, we get capital 𝐻 of 𝑦 plus a constant, which we’ll call 𝑐 one, is equal to capital 𝐺 of π‘₯ plus a constant, which we’ll call 𝑐 two. And this is when capital 𝐻 and capital 𝐺 are antiderivatives of a lowercase β„Ž and lowercase 𝑔, respectively. We should now pay attention to these two 𝑐 terms, which are constants of integration. Given that these are constants, we can simplify first by subtracting 𝑐 one from both sides and then simply by redefining 𝑐 two minus 𝑐 one to be a new constant, which we’ll call 𝑐 three.

At this stage, we would have found the general solution to our differential equation. It is likely that this will be an implicit solution. Remember, an explicit solution is one with 𝑦 on one side of the equation and some function of π‘₯ on the other side. An implicit solution does not take this form. Both forms of solution are valid. But we often try to express our answer in terms of an explicit solution if possible. Now, here, we’ve said that the general solution has been found because the constant 𝑐 three can take any value. And all of these will be solutions to our differential equation.

Let’s imagine the very simple case where 𝐻 of 𝑦 is just 𝑦 and 𝐺 of π‘₯ is just π‘₯. Our equation then becomes 𝑦 equals π‘₯ plus our constant 𝑐 three. Of course, this is just the equation of a straight line. Since 𝑐 three can take any value, if we took it to be equal to zero, we’d have the straight line 𝑦 equals π‘₯. When 𝑐 three is equal to minus one, we’d have the line 𝑦 equals π‘₯ minus one. We could even take something like 𝑐 three equals πœ‹, which would give us 𝑦 equals π‘₯ plus πœ‹. All of these lines and their corresponding equations are specific solutions to some differential equation. We found these using the general solution to that differential equation. And we’ve already shown the method for finding a general solution to a differential equation.

Okay, so without just arbitrarily picking numbers for our constant, how do we get from the general solution to the specific solution? As per the title of this video, we’ll be using an initial value, sometimes called an initial condition. Let’s see an example of this using our general solution 𝑦 equals π‘₯ plus some constant. An example of an initial value that we might be given would be 𝑦 of two is equal to one. This tells us that when π‘₯ is equal to two, 𝑦 is equal to one. It, therefore, follows that the point two, one lies on the line of the correct specific solution that we’re looking for.

Algebraically, we can find the equation of the correct specific solution by substituting in the known values into the general solution. When π‘₯ equals two, 𝑦 equals one. Of course, a simple rearrangement gives us that the constant 𝑐 three is equal to negative one. Substituting this back into our general solution, we get the equation 𝑦 equals π‘₯ minus one. A visual representation of what we’ve done here is that we looked at the point two, one. And we found the equation of the line for which it lies on. This line is the specific solution which matches the initial value we were given.

Now that we understand our method, let’s take a look at an example.

Find the equation of the curve which passes through the point negative eight, one given that the gradient of the tangent at any point is equal to two times the square of the 𝑦-coordinate.

Let’s first interpret the information given by this question. We’re told about the gradient to the tangent of a curve at any point. This gradient can be represented by d𝑦 by dπ‘₯. We’re told that this is equal to two times the square of the 𝑦-coordinate. So in other words, is equal to two 𝑦 squared. Here, we recognize that we’ve been given a differential equation. Not only that, we’ve been given a separable differential equation. This is one that can be expressed in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. In our case, we’ll be taking 𝑔 of π‘₯ to be equal to the constant two and 𝑓 of 𝑦 to be equal to 𝑦 squared.

Given an equation of this form, we can find the general solution in the following way. We first divide both sides of our equation by 𝑓 of 𝑦. In our case, this is 𝑦 squared. This gives us that one over 𝑦 squared d𝑦 by dπ‘₯ is equal to two. Next, we can integrate both sides of our equation with respect to π‘₯. Now, the chain rule tells us that d𝑦 by dπ‘₯ dπ‘₯ is equal to d𝑦. This gives us that the integral of one over 𝑦 squared with respect to 𝑦 is equal to the integral of two with respect to π‘₯. As a side note, you often see methods where before the integration, d𝑦 by dπ‘₯ is treated somewhat like a fraction. Both sides are multiplied by dπ‘₯.

And the equivalent statement is obtained: one over 𝑦 squared d𝑦 is equal to two dπ‘₯. Both sides of the equation are then integrated. And we reach the same state as we did before, but with less working. Always remember that this is just a shorthand for the more rigorous steps that we’ve shown here. Returning to our method, one over 𝑦 squared can be expressed as 𝑦 to the power of negative two. You might be more comfortable performing the integration in this form to get negative one over 𝑦 plus the constant of integration 𝑐 one is equal to two π‘₯ plus some other constant of integration, 𝑐 two.

Usually, we don’t worry about putting a constant on both sides of this equation because we just define a new constant, which in this case would be 𝑐 two minus 𝑐 one. And this could just go on one side of the equation. Now, at this stage, we have found the general solution to our differential equation. And it’s in implicit form, which means it’s not in the form where 𝑦 is equal to some function of π‘₯. We found the general solution for all lines where the gradient of the tangent at any point is equal to two times the square of the 𝑦-coordinate.

Now, our next thought might be to convert our implicit solution into an explicit solution. But as it turns out, this isn’t necessary. This is because the general solution is not our final destination. Rather, we are seeking a particular solution. We want the line which passes through the point negative eight, one. This information is an initial value. And we’ll be finding the particular solution which matches in the following way. We know that if negative eight, one lies on the line, when π‘₯ is equal to negative eight, 𝑦 must be equal to one. We can substitute these known values into the equation for our general solution. We then perform some simplification. And we find that our constant 𝑐 three is equal to 15.

We can now substitute this value into our general solution which we’ll turn it into a particular solution. Negative one over 𝑦 is equal to two π‘₯ plus 15. Now, if we wanted, we could convert this to an explicit solution first by multiplying by negative one and then raising both sides to the power of negative one. To recap, we used the information given in the question to form a separable differential equation. The particular solution to this which passes through the point negative eight, one is 𝑦 equals negative one over two π‘₯ plus 15. And this is the answer to our question.

Let’s now take a look at another example to practise this form of question.

Find the equation of the curve that passes through the point zero, negative one given d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four divided by four 𝑦 plus 13.

For this question, we first noticed that we have been given a separable differential equation. This is an equation that can be written in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. Solving this separable differential equation will give us a general solution, which will include a constant 𝑐. Now, the question has also given us an initial value, which is the fact that the curve should pass through the point zero, negative one. We’ll be able to use this information to narrow down the general solution that we find into a specific solution. For now, let’s just work on the general solution.

For now, we’ll put aside our initial value and just work towards the general solution. To find this, we’ll first express our differential equation in the following form, where we have some function of 𝑦 d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯. For our equation, we can do this by multiplying both sides by four 𝑦 plus 13. This gives us that four 𝑦 plus 13 d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four. The next thing we can do is to treat d𝑦 by dπ‘₯ somewhat like a fraction. If we pretend we’re multiplying both sides by dπ‘₯, we obtain an equivalent statement four 𝑦 plus 13 d𝑦 is equal to negative six π‘₯ minus four dπ‘₯.

The next step we take is to then integrate both sides of our equation, giving us an integral with respect to 𝑦 on the left-hand side and an integral with respect to π‘₯ on the right-hand side. Now, you should always remember that what we have done here is essentially a trick. d𝑦 by dπ‘₯ is not actually a fraction and cannot always be treated like one. However, we’ve done so here to save us from working since it gives us an equivalent answer. Going back to our method, we now perform our integrations. Doing so gives us the following.

Now, it’s worth noting that both of our integrals would actually have given us a constant. However, we can simply combine these two constants into one term. Instead of 𝑐 one and 𝑐 two, we simply have a 𝑐, which is equal to 𝑐 one minus 𝑐 two. Again, this leads to less unnecessary work for us. After some small simplifications, we have arrived at the general solution to our differential equation. This means that we have found an equation which describes all the lines for which d𝑦 by dπ‘₯ is equal to negative six π‘₯ minus four divided by four 𝑦 plus 13.

However, we need to find which one of these lines passes through the point that we’ve been given: zero, negative one. This initial value that we have been given will allow us to find that particular solution. We know that if the line passes through the point zero, negative one that when π‘₯ is equal to zero, 𝑦 must be equal to negative one. We can, therefore, take our general solution, substitute in 𝑦 equals negative one and π‘₯ equals zero, and solve this to find the value of our constant 𝑐. With a little bit of working, we find that the value of 𝑐 here is negative 11. We can now substitute this value of 𝑐 back into our general solution to find the specific solution. This is the specific solution to the differential equation which passes through the point zero, negative one, which is our initial value.

Now, you might notice that the solution that we have been given is implicit, which means it’s not in the form 𝑦 equals some function of only π‘₯. Usually, it’s a good idea to try and express your solution in explicit form. However, in this case, you should not worry. It is not actually possible to give the answer to this problem in the form of a single explicit equation since we have a multiple powers of 𝑦 on the left-hand side. Again, this is no cause for concern, since the implicit form is a perfectly valid solution. And we have indeed answered our question.

Let us now take a look at one final example of an initial value problem.

Find the solution of the differential equation d𝑦 by dπ‘₯ plus nine 𝑦 is equal to 63 given that 𝑦 nought is equal to eight.

For this type of question, the first thing we should do is to check whether we have been given a separable differential equation. This is a differential equation that can be written in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. An equivalent statement to this would be one over 𝑓 of 𝑦 d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯. We might also decide to define another function β„Ž of 𝑦 to make notations slightly easier, where β„Ž of 𝑦 is equal to one over 𝑓 of 𝑦.

Now, here we should be a little bit careful. Just because our equation has all of the 𝑦 terms on the left does not mean it’s already in this form. This is because we have an addition here. Consider the following side example of the differential equation d𝑦 by dπ‘₯ plus 𝑦 is equal to two π‘₯. Although it might look like this is a separable differential equation because we have a 𝑦 on the left and an π‘₯ on the right. In actual fact, no matter how much we try, we cannot express it in the correct way. This example illustrates that some equations which look like they should be separable in fact are not. Luckily, we are not dealing with one of these cases. However, our equation does require a little bit of manipulation before we proceed.

We take our equation and we subtract nine 𝑦 from both sides. We can then factorize the right-hand side since both terms have a common factor of nine. We then divide both sides by seven minus 𝑦. This gives us the following equation which does indeed match the form that we’ve shown here. Okay, we’re now gonna use a little bit of a trick. We can treat d𝑦 by dπ‘₯ somewhat like a fraction. Doing so allows us to reach a different form of equation: one over seven minus 𝑦 d𝑦 is equal to nine dπ‘₯. From this form, we can then integrate both sides of our equation. This gives us the negative of the natural logarithm of the absolute value of seven minus 𝑦 is equal to nine π‘₯ plus 𝑐, where 𝑐 is a constant. Of course, both of our integrals would have given us a constant of integration. However, we’ve chosen to combine both of these on the right-hand side into the constant, which we’ve just called 𝑐.

Okay, we’ve now found the general solution to our differential equation. However, this solution is in an implicit form. An explicit solution would be one in the form 𝑦 equals some function of π‘₯. We’ve chosen to use 𝑒 here to avoid confusion with this 𝑓 in our definition. To make things easier to manage, let’s work towards converting our general implicit solution to a general explicit solution. The first thing we can do is multiply both sides by negative one. Since we haven’t yet defined our constant, it doesn’t matter whether we have a plus 𝑐 or a minus 𝑐. So perhaps, we’ll just leave it as a plus 𝑐. Let’s clear some room to continue.

The next thing we do is take the exponential of both sides, which means raising 𝑒 to the power of both sides of our equation. Since this is the inverse of the natural algorithm on the left-hand side of the equation, we’re simply left with seven minus 𝑦. Next, we can simplify by subtracting seven from both sides and then multiplying by negative one. This gives us that 𝑦 is equal to negative 𝑒 to the power of negative nine π‘₯ plus 𝑐 add seven. At this stage, we might recognize that 𝑒 to the power of negative nine π‘₯ plus 𝑐 is equal to 𝑒 to the power of negative nine π‘₯ multiplied by 𝑒 to the power of 𝑐. Doing this gives us another way to simplify.

Together here, we have negative 𝑒 to the power of 𝑐. But since 𝑐 is currently an undefined constant, we can redefine a new constant. Let’s say capital 𝐢, which is equal to negative 𝑒 to the power of our previous lowercase 𝑐. Doing so allows us to say that 𝑦 is equal to capital 𝐢 times 𝑒 to the power of negative nine π‘₯ plus seven. We have now found a general implicit solution to our differential equation. At this stage, we’re ready to use the remaining information given by the question. The question has given us that 𝑦 of nought is equal to eight this information is our initial value and we need to find the particular solution to the differential equation for which this is true.

From this statement, we know that when π‘₯ is equal to zero, 𝑦 must be equal to eight. Another way of saying this would be that the curve which represents our particular solution will pass through the point zero, eight. Okay, we can substitute our known values into the general solution to find the value of capital 𝐢 for our particular solution. Since 𝑒 to the power of nought is equal to one, we have that eight is equal to 𝐢 times one add seven. Therefore, our constant capital 𝐢 is equal to one. Substituting this value of 𝐢 back into our general solution gives us the particular solution to our differential equation, which matches the initial value given by the question. The equation that represents our particular solution is that 𝑦 is equal to 𝑒 to the power of negative nine π‘₯ plus seven.

Okay, with our final example done, let’s finish off the video by going through some key points. A separable differential equation can be expressed in the form d𝑦 by dπ‘₯ is equal to 𝑔 of π‘₯ multiplied by 𝑓 of 𝑦. There are also other ways in which this can be expressed, which may be more helpful depending on the problem. The general solution to a separable differential equation can be found using integration techniques. Within this method, you often see d𝑦 by dπ‘₯ treated somewhat like a fraction. But remember, this is just a trick which saves your working. The general solution to a differential equation will include a constant 𝑐.

An initial value, for example, 𝑦 of π‘Ž is equal to 𝑏, can be used to find a particular solution to the differential equation. This can be done by substituting in π‘₯ equals π‘Ž and 𝑦 equals 𝑏 into your general solution and then solving to find the constant 𝑐. Putting this value of 𝑐 back into your general solution will give rise to a particular solution, specifically one that matches the initial condition that you used.

As a final side note, solutions can either be explicit or implicit. Explicit solutions are of the form 𝑦 equals some function of π‘₯. Implicit solutions are not of this form. It’s usually a good idea to try and give an explicit solution to a problem. But sometimes this is impossible. So don’t worry too much if you cannot do this.

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