# Video: Indefinite Integrals and Initial Value Problems

In this video, we will learn how to use integration to find particular solutions to initial value problems involving differential equations of the form d𝑦/d𝑥 = 𝑓(𝑥).

14:33

### Video Transcript

In this video, we’ll learn how to use integration to find particular solutions to initial value problems involving differential equations of the form d𝑦 by d𝑥 equals 𝑓 of 𝑥. A differential equation is an equation that contains derivatives or differentials. These differentials are of the form 𝑦 prime equals 𝑓 of 𝑥 or d𝑦 by d𝑥 equals some function in 𝑥. We’ll begin by looking at how integrals can help us to find a general solution to these equations before considering what the introduction of an initial value does to these solutions.

Let’s begin by recapping how integration can help us with differential equations. Here is a simple differential equation. d𝑦 by d𝑥 equals three 𝑥 squared plus four. In other words, there’s some function 𝑦 in terms of 𝑥 that when we differentiate it with respect to 𝑥, we end up with its derivative being three 𝑥 squared plus four. But then, we recall that integration and differentiation are the reverse processes of one another. In other words, if we integrate this function, its derivative with respect to 𝑥, we’ll end up with the general solution to the differential equation. We obtain the original function 𝑦 involving some constant of integration 𝑐.

In this case then, 𝑦 is going to be equal to the indefinite integral of three 𝑥 squared plus four with respect to 𝑥. We then recall that to integrate terms in a polynomial, we add one to the power and then divide by this new number. Assuming that exponent is not equal to negative one. So the integral of three 𝑥 squared is three 𝑥 cubed divided by three. And the integral of four is four times 𝑥 to the power of one divided by one. And of course, since we’re working with an indefinite integral, we have that constant of integration 𝑐. Simplifying, and we see that 𝑦 is equal to 𝑥 cubed plus four 𝑥 plus 𝑐.

This is the general solution to our differential equation. But what about its particular solution? In other words, how do we work out the value of 𝑐? Well, we need an initial value. Let’s have a look at a simple initial value problem.

Find a particular solution for the following differential equation for which 𝑦 of zero equals 12. d𝑦 by d𝑥 equals eight 𝑥 plus three.

Remember, we can find a general solution for the differential equation d𝑦 by d𝑥 equals eight 𝑥 plus three by first performing the reverse process for differentiation. We’re going to integrate our function for d𝑦 by d𝑥 with respect to 𝑥. So 𝑦 is equal to the indefinite integral of d𝑦 by d𝑥 with respect to 𝑥. Or 𝑦 equals the integral of eight 𝑥 plus three with respect to 𝑥. We then recall that the integral of a polynomial term of the form 𝑎𝑥 to the 𝑛th power, where 𝑎 and 𝑛 are real constants and 𝑛 is not equal to negative one, is 𝑎𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus some constant of integration 𝑐.

Essentially, we add one to the exponent and then divide by that new number. This means the integral of eight 𝑥 is eight 𝑥 squared over two. Then the integral of three is three 𝑥. And, of course, since this is an indefinite integral, we end up with that constant of integration 𝑐. So we obtained the general solution of our differential equation to be 𝑦 equals four 𝑥 squared plus three 𝑥 plus 𝑐.

Now, let’s refer back to the question. We’re told that 𝑦 of zero equals 12. In other words, at some point in the equation 𝑦 equals 𝑓 of 𝑥, when 𝑥 is equal to zero, 𝑦 is equal to 12. Let’s substitute these values into our equation and see what happens. We obtain 12 equals four times zero squared plus three times zero plus 𝑐. Well, this whole equation simplifies to 12 equals 𝑐. And that’s great because now we know the value of our constant. And we can, therefore, say that 𝑦 equals four 𝑥 squared plus three 𝑥 plus 12.

This is known as the particular solution. And 𝑦 of zero equals 12 is an initial value. These are sometimes called initial value problems as they allow us to find a particular solution by giving us an output of a specific 𝑥-value.

Let’s now have a look at another example.

Find the solution for the following differential equation for 𝑦 of zero equals one. d𝑦 by d𝑥 minus 𝑥 minus 𝑥 squared equals zero.

In this question, we have a differential equation and an initial value. That is, when 𝑥 is equal to zero, 𝑦 is equal to one. Now, before we can look to solve this differential equation, we’re going to need to perform an intermediate step. We’re going to rearrange and make d𝑦 by d𝑥 the subject. Now, it’s not particularly difficult to achieve this. We add 𝑥 and 𝑥 squared to both sides of our differential equation to find that d𝑦 by d𝑥 equals 𝑥 squared plus 𝑥. Now, remember, we can find a general solution for the differential equation d𝑦 by d𝑥 equals 𝑥 squared plus 𝑥 by performing the reverse process to differentiation. We’re going to integrate with respect to 𝑥. So 𝑦 will be equal to the integral of d𝑦 by d𝑥 with respect to 𝑥 or the integral of 𝑥 squared plus 𝑥 with respect to 𝑥.

Remember, to integrate a polynomial term, we add one to the exponent and then divide by that new value. So the integral of 𝑥 squared is 𝑥 cubed over three. We then see that the integral of 𝑥 is 𝑥 squared over two. And since we’re working with indefinite integrals, we know we need that constant of integration 𝑐. So 𝑦 is equal to 𝑥 cubed over three plus 𝑥 squared over two plus 𝑐. So we have the general solution to our differential equation. But we haven’t yet used the information that 𝑦 of zero equals one. We’re going to substitute 𝑥 equals zero and 𝑦 equals one into this general solution. When we do, we see that one equals zero cubed over three plus zero squared over two plus 𝑐. And this whole equation simplifies to one equals 𝑐. So we found that 𝑐 equals one. And the particular solution to our differential equation is 𝑦 equals 𝑥 cubed over three plus 𝑥 squared over two plus one.

Now, it’s important to realize that we can also apply this procedure to more complicated examples.

Find the function 𝑓, if 𝑓 prime of 𝑡 equals two sec 𝑡 times tan 𝑡 plus four sec 𝑡, when 𝑡 is greater than negative 𝜋 by two and less than 𝜋 by two and 𝑓 of negative 𝜋 by three is negative two.

In this question, we have a differential equation. Remember, this is just an equation involving derivatives. Here, that’s 𝑓 prime of 𝑡. And we’re also given an initial value. That is, when 𝑡 is equal to negative 𝜋 by three, 𝑓 of 𝑡 is equal to negative two. Now, we’re looking to work out what the function 𝑓 is. And so, we recall that integration and differentiation are reverse processes. So our original function 𝑓 of 𝑡 will be the integral of 𝑓 prime of 𝑡 with respect to 𝑡. That’s the integral of two sec of 𝑡 times tan 𝑡 plus four sec of 𝑡 with respect to 𝑡.

Now, this doesn’t look very nice. Let’s first distribute the parentheses of our integrand. When we do, we obtain our function 𝑓 of 𝑡 to be equal to the integral of two sec 𝑡 tan 𝑡 plus eight sec squared 𝑡 with respect to 𝑡. We can simplify this a little bit by recalling that the integral of the sum of two functions is equal to the sum of the integral of each of those functions. So 𝑓 of 𝑡 is equal to the integral of two sec 𝑡 tan 𝑡 with respect to 𝑡 plus the integral of eight sec squared 𝑡 with respect to 𝑡.

Now, these might look really nasty. But let’s recall some derivatives. We know that the derivative of sec of 𝑡 with respect to 𝑡 is tan 𝑡 sec of 𝑡. And we know that the derivative of tan 𝑡 with respect to 𝑡 is sec squared 𝑡. This means the antiderivative of tan of 𝑡 sec of 𝑡 must be sec of 𝑡. And the antiderivative of sec squared 𝑡 must be tan 𝑡. And so, the integral of two sec of 𝑡 tan 𝑡 must be two sec 𝑡. And, of course, it’s an indefinite integral. So we’ll add a constant of integration 𝐴. Then the integral of eight sec squared 𝑡 must be eight tan 𝑡. And we’ll add a second constant of integration 𝐵. Combining these two constants of integration, we see that 𝑓 of 𝑡 is equal to two sec 𝑡 plus eight tan 𝑡 plus 𝐶.

This is the general solution to our differential equation. But we haven’t yet used the fact that when 𝑡 is equal to negative 𝜋 by three, our function is equal to negative two. So let’s substitute these values in. And when we do, we obtain negative two to be equal to two sec of negative 𝜋 by three plus eight tan of negative 𝜋 by three plus 𝐶. If we recall that sec of negative 𝜋 by three is one over cos of negative 𝜋 by three. Then two sec of negative 𝜋 by three must be two over cos of negative 𝜋 by three, which gives us four. Then eight tan of negative 𝜋 by three is negative eight root three. So negative two is equal to four minus eight root three plus 𝐶.

Let’s subtract four from both sides of this equation and add eight root three to both sides. And we see that 𝐶 is equal to eight root three minus six. And we’ve obtained the particular solution to our differential equation to be 𝑓 of 𝑡 equals two sec 𝑡 plus eight tan 𝑡 plus eight root three minus six.

Now, the real power of this process is in contextual examples. Let’s have a look at one.

The acceleration, 𝑎 meters per square second, of a particle at time 𝑡 seconds is given by the equation 𝑎 equals negative three sin of 4𝑡 for 𝑡 is greater than or equal to zero. The initial velocity of the particle is three meters per second and the initial displacement of the particle is negative two meters. Find an equation for the displacement, 𝑠, of the particle at time 𝑡 seconds.

Let’s begin by recalling the relationship between acceleration, velocity, and displacement. Let’s say we have a function for displacement, 𝑠, in terms of 𝑡. The velocity of an object is its rate of change of displacement with respect to time. And, of course, when we consider rate of change, we’re actually thinking about derivatives. So we can say that 𝑣 is equal to d𝑠 by d𝑡. Similarly, acceleration is defined as rate of change of velocity with respect to time. So acceleration is d𝑣 by d𝑡. And since 𝑣 itself is the first derivative of 𝑠 with respect to time, we can say that acceleration could also be known as the second derivative of displacement with respect to time.

Now, in this question, we’ve been given a function for acceleration in terms of time. To be able to find a function for displacement, we’re going to need to do the reverse process for differentiation. We’re going to integrate our function for acceleration with respect to time. That will give us a function for velocity. Then, we’re going to integrate once more to find our function for displacement. So let’s begin by integrating our function for acceleration. We know that 𝑣 must be equal to the integral of negative three sin of four 𝑡 with respect to 𝑡. Well, remember, we can take out any constant factors and focus on integrating the function with respect to 𝑡. So we can say that the integral of negative three sin of four 𝑡 with respect to 𝑡 is negative three times the integral of sin of four 𝑡 with respect to 𝑡.

Then, we recall the standard result for the integral of sin of 𝑎𝑥 with respect 𝑥. It’s negative one over 𝑎 cos 𝑎𝑥. And so, this means the integral of sin of four 𝑡 must be negative one-quarter cos of four 𝑡. And, of course, we need that constant of integration here. I’ve called that 𝐴. Distributing our parentheses, we therefore see that 𝑣 is equal to three-quarters cos of four 𝑡 plus 𝐵. And the reason our constant of integration has become 𝐵 is because we’ve changed it. We’ve multiplied it by negative three.

We’re told that the initial velocity of the particle is three meters per second. In other words, when 𝑡 is equal to zero, 𝑣 must be equal to three. So let’s substitute these values into our function for velocity. That gives us three equals three-quarters cos of zero plus 𝐵. And since cos of zero is one, we see that three is equal to three-quarters plus 𝐵. To solve for 𝐵, we subtract three-quarters from both sides of the equation. And we find that 𝐵 is equal to nine-quarters. And so, we’ve found that velocity is equal to three-quarters cos of four 𝑡 plus nine-quarters.

Our next step is to integrate our function for velocity to find a general function for displacement. That’s the indefinite integral of three-quarters cos of four 𝑡 plus nine-quarters with respect to 𝑡. This time, we recall that the integral of cos of 𝑎𝑥 is one over 𝑎 sin of 𝑎𝑥. And so, the integral of three-quarters cos of four 𝑡 must be three-quarters times a quarter sin of four 𝑡. Then the integral of nine-quarters is nine-quarters 𝑡. So we have a general function for displacement. It’s three-quarters times one-quarter sin of four 𝑡 plus nine-quarters 𝑡 plus, of course, that constant of integration, 𝑐. This simplifies to three sixteenths sin of four 𝑡 plus nine-quarters of 𝑡 plus 𝑐.

There’s one piece of information we’ve not yet used. The initial displacement of the particle is negative two meters. So when 𝑡 is equal to zero, not only is the velocity three meters per second, but 𝑠 is equal to negative two. Let’s substitute these values into our equation. That’s negative two equals three sixteenths sin of zero plus nine-quarters of zero plus 𝑐. And sin of zero is zero, and nine-quarters of zero is zero. So we see that negative two is equal to 𝑐. And so, our very final step is to replace 𝑐 with negative two in our equation for the displacement. And we find the equation with the displacement 𝑠 of the particle at time 𝑡 seconds is three sixteenths sin of four 𝑡 plus nine-quarters 𝑡 minus two.

In this video, we’ve seen that we can use integration to find general solutions of differential equations. We also saw that we can use initial values to evaluate the constant. And this is called the particular solution to the differential equation.