Initial Value Problems - Online Tutor, Practice Problems & Exam Prep

Initial Value Problems

At this point in the course, there are a couple of things we should be quite familiar with. We should be familiar with the process of taking a derivative, and we should know in general what this notation looks like, and we should also be familiar with the techniques of integration. We've learned about integrals a lot in recent videos, and what we're going to be doing in this video is combining these ideas of derivatives and integrals into something known as an initial value problem. So without further ado, let's see how we can solve these types of problems, and if you're familiar with the techniques in general for both integrals and derivatives, I think you'll find this pretty straightforward. Now to understand how to solve initial value problems, what you first need to do is know what a differential equation is.

A differential equation is an equation involving some unknown function as well as its derivative. So an example of a differential equation would be what we have right down here. So notice how we have this derivative, and we have the, what the unknown function, what its derivative is equal to, meaning we have a differential equation. Now in this problem specifically, we're told given this differential equation right here, and this condition over here, y of 1 is equal to negative one, we're asked to find y of x. Now whenever you have a situation where you're given a differential equation, and it's combined with some initial condition, like we see right here, that means that you're dealing with an initial value problem.

Now the initial condition that we have, this is basically just giving us a point. Right? Y of 1 equals negative 1 but it's giving us a point of the original function that we don't have yet. We have its derivative, so we're going to need to figure out what the original function is first, and then use this point to solve the problem. So first off, we're going to deal with the differential equation that we have down here.

And the way that I can figure out what this function y of x is, is by using integrals. Because notice that we have a derivative given to us, so if I use an integral, that's the reverse process for taking a derivative. So integrating dydx, well, that's just going to get me to my original function y. And if I integrate the other side of this equation as well, well, that's going to allow me to see what this function y of x is. Now I can use the sum and difference rule here and integrate these terms separately.

The integral of 3x² using the power rule is going to give me 3x³ over 3. The integral of negative 4 is going to be negative 4x, and then we're going to have plus the constant of integration C. Now at this point, I can cancel the threes right here, meaning all I'm going to have is x³. That's going to be minus 4x plus C. So this is what we get when we do this integral of the right side.

Now this equation that we just figured out here, this is what's called the general solution. Now this is actually not the final solution to the problem, and here's the reason why. We have a plus C right here. We need to know what this constant is if we want to find this final equation for y of x. We can't just leave it as a plus C because this is only the general solution.

So how exactly could we find what this plus C is? Well, notice we're also given an initial condition, and that initial condition is applied when we have the original function. We just found the original function using this integration, so I can apply this initial condition that's given to us to solve this problem. So we're told that y of 1 is equal to negative 1. So that means you're going to have y of 1, meaning I need to take every place that I see x and replace it with 1.

So we're going to have 1³ minus 4 times 1 plus C. Now we're told that this whole thing is equal to negative one. So that means this whole thing is equal to negative one. 1³ is 1, and we have 4 times 1 which is 4 plus C. Now from here, what I can do is I can go ahead and subtract 4 from the one here.

So negative one is equal to negative 3 plus C. And then what I'll do is add this 3 on both sides of the equation here. That's going to get the threes to cancel there. I'll add the 3 on both sides, giving me that C is equal to negative one plus 3 which is equal to 2. So that means that our constant C is equal to 2.

So because we have that C is equal to 2, that means that we can take the C up in this equation and replace it with 2. And this is going to give us the particular solution to our problem when we go ahead and apply this initial condition that we have. So taking this C up here, placing it with 2, will give us that y of x is equal to x³ minus 4x plus 2. So this right here would be the final solution to the problem. As you can see, it's pretty straightforward.

We're really just applying the techniques we've already learned about for specifically integration, and also our conceptual understanding of derivatives to solve problems like this. Now something else that I'll mention is it's possible you'll see these differential equations or these initial value problems where you have situations with higher-order derivatives. So you might see a second derivative or a third derivative. But whenever this happens, you're just going to go ahead and apply the same process multiple times for solving these types of problems. So in this case, we had to integrate once that gave us y of x.

But if we had a second derivative, we'd have to integrate once to get y prime of x. We would have to integrate again to get y of x, and then so on and so forth. So you just apply the same process multiple times, and typically, you'll be given more than one initial condition if you have more than one derivative. So that's how you solve these types of problems. Hope you all found this video helpful, and let's go ahead and try getting some more practice with some of these more complicated situations to see how we can solve all kinds of initial value problems.

See you in the next video.

Solve the following initial value problem:

$\frac{dy}{dx}=2x-5$; $y\left(0\right)=4$

• A. $y\left(x\right)=2x^2-5x$
• B. $y\left(x\right)=x^2-5x$
• C. $y\left(x\right)=2x^2-5x+4$
• D. $y\left(x\right)=x^2-5x+4$

Using the acceleration function below, find the velocity function, if the velocity is v = 5 at time t = 2.

$a\left(t\right)=-20$

• A. $v\left(t\right)=-20t+10$
• B. $v\left(t\right)=-20t+5$
• C. $v\left(t\right)=-20t+45$
• D. $v\left(t\right)=-20t-35$

Find the function $f\left(x\right)$ that satisfies the following differential equation.

$f^{\prime\prime}\left(x\right)=3x^2$; $f^{\prime}\left(0\right)=1$; $f\left(1\right)=3$

• A. $f\left(x\right)=3x^4+x-1$
• B. $f\left(x\right)=\frac{x^4}{4}+x+\frac74$
• C. $f\left(x\right)=3x^4+x+\frac74$
• D. $f\left(x\right)=\frac{x^4}{4}+x-1$

Initial Value Problems Example 1

So here in this example, we are told to solve the initial value problem given by this function right here. So we have this differential equation, and notice we are given the initial conditions over here. Now this can be a bit of a tricky initial value problem since we have 2 derivatives that are being taken right here. But it turns out that the process for solving these differential equations where you have multiple derivatives is the exact same process we've seen when dealing with only one derivative. We just have to apply multiple conditions and do this process multiple times.

So let's go ahead and get started. Now we have this derivative right here. It says that the second derivative of y with respect to x is equal to this function, x3 multiplied by 4, and then we're going to have minus 10. So that's the function that we have. And notice that this first initial condition here says that y prime of 1 is equal to negative 2.

Now we also have that y of 0 is equal to 3, but I'm going to ignore that initial condition just for now. I'm going to focus on these 2 at the moment. We'll get to this later, but let's just start with these 2. So what I can recognize is that if I take this function and integrate it, I'm going to be doing the reverse process for a derivative. So that's going to get me from 2 derivatives down to just one derivative, because, again, I'm in essence reversing this derivative.

But to do this, I would need to integrate this entire right side of the equation, and I can do that by applying the rules for integrals that I know. So see that we have a difference between x3 multiplied by 4 and 10, which means that we're going to have the integral of 4x3 individually minus the integral of 10. Now from here, I can take this constant of 4 and pull it outside of the integral, so that the integral of x3 with respect to x minus the integral of 10, which I'll leave inside the integral here since that 10 is by itself. So what I'm going to do is use the power rule on this x cube here to evaluate this integral, meaning we're going to get 4, and it's gonna be x4 over 4. Just adding 1 to this exponent and dividing by that new exponent of 4.

And then we're going to have minus the integral of 10 with respect to x, which is just 10x, And then don't forget the constant of integration c. Now notice here that this 4 will cancel with that 4, leaving us with x4 minus 10x plus c as my general solution for dy over dx. Now notice that we're told that y prime of 1 is equal to negative 2, and this is the general solution we've got. Dy over dx can also be written as y prime of x. Because if you think about it, that's just the first derivative of this function y.

So we have x4 minus 10x plus c. And what I'm going to do is take this initial condition that I have up here. I'm gonna see if I can apply it to what we have down here. So doing that, we have y prime of 1 is equal to negative 2. I'm gonna go ahead and move that down here so I can do this work.

So we're going to have that y prime of 1 is negative 2. So we'll have y prime of 1, meaning I need to take every place that I see x here and replace it with 1. So 1 to the 4th power minus 10 times 1 plus the constant c. Now y prime of 1 is negative 2. And then we have 1 to the 4th power, which is 1, minus 10 times 1, which is 10, plus c.

Now 1 minus 10, that's going to give us negative 9 plus c. Then I just need to add this 9 on both sides of the equations. That's going to get the nines to cancel on the right side here, leaving me with just c. Now if c is equal to 9 plus negative 2, which is positive 7. So this is our solution for c.

So going back up here to the general solution that we calculated, I can knock out that c and replace it with 7 since that's the specific answer that we get for this function y prime of x. So we've gone ahead and found y prime of x, but we're not quite done yet because we're not trying to find y prime of x. We're actually trying to find y of x because we're, in essence, trying to get this down to just the function y of x. And notice that we have another condition that we can apply up here to do that. So what I'm going to do is take this function I just calculated, and I'm going to write it up here.

So we have that y prime of x is equal to x4 minus 10x plus 7. I'm just rewriting this solution that we got up here. And what I'm going to do is go through the same process again. So we had 2 derivatives taken here for this entire function, and then we just use this initial condition. Now I'm going to jump to that initial condition.

So we have y of 0 equals 3, but I can't apply that yet because I first need to figure out what y of x is. Well, if we have y prime of x, I can integrate to give me y of x. If I integrate this right side of the equation, well, I can integrate each one of these terms where we have the sums and differences separately. So the integral of x4 with respect to x minus the integral of 10x with respect to x plus the integral of 7 with respect to x. Now using the power rule, the integral of x4 is going to give me x5 over 5.

Now this is going to be minus. And by the way, this integral could also be written as 10 times the integral of x with respect to x. It won't have plus, and then the integral of 7 with respect to x is 7x. Then we're gonna have plus the constant of integration c. That should also deal with this integral right here.

I can use the power rule on this x here, which will give us x5 over 5 minus, and then we'll have 10 times this integral here, which is going to be x squared over 2. Again, just using the power rule where we add 1 to the exponent and then divide by the exponent. So that's why we get this x squared over 2. And then, of course, we already figured out the 7x, and we figured out the plus c. Now from here, I can reduce distractions.

We have x5 over 5 minus, and then 10 divided by 2 is going to give us 5x squared, just simplifying and cleaning things up. We'll have plus 7x plus c, and that's gonna be our y of x. Now from here, notice there's another condition that I can apply, which is that y of 0 is equal to 3. So what I'm going to do is apply that condition to the function that we just calculated. So doing this, I'm gonna make some space for myself because we already solved this portion of the problem.

So I'll move all of this over to the left. And now I'm just gonna go ahead and go through the process here to figure out what c is. Then to solve for c, well, I'm going to take every place that I see x and replace it with 0 in this function y. You're going to have that y of 0 is equal to 0 to the 5th power over 5 minus 5 times 0 squared plus 7 times 0 plus c. Now notice each of these terms are going to go to 0 since we're multiplying by 0, meaning that we're just going to have that y of 0 is equal to 0 plus c.

And we can see here that y of 0 is equal to 3, so that means 3 is equal to 0 plus c, which is just c. So c is equal to 3. So if you get out the constant is 3, meaning going back up to this general solution here, we can replace that constant c with 3. So this right here is gonna be our solution for y of x, and that is so you can solve this situation or this initial value problem. So as you can see, it gets a little bit more tedious when we're dealing with 2 derivatives rather than just 1, but it still is the same process where you're applying the initial conditions and taking in essence the reverse process or antiderivative to figure out what exactly that function is.

So I hope you found this video helpful, and let's get some more practice.

Initial Value Problems Example 2

An object moving along a line is described by the acceleration function a(t). Find the position function s(t), given the initial velocity v(0) = 9, and the initial position s(0) = 0. So how can we go about solving a problem like this? Well, this problem is quite tricky because we are dealing with an initial value problem where we have this conceptual story, and we need to, in essence, kind of work backwards here from our derivatives to understand what's happening. If you pay close attention, I'm going to walk you through how to solve a problem like this.

If you're able to solve this, then you should be able to handle the more complicated problems that you see with initial values. So let's just go ahead and get right into things. Now what I'm first going to do is figure out what exactly I'm looking for here. And I can see I'm looking for the position function s(t). That's the main goal of this problem.

Now what I've given is the acceleration function. So how could I use this acceleration function to find position? Well, I need to think about how position and acceleration are related. And I know that the first derivative of position is going to give me velocity. I know that the second derivative of position is the first derivative of velocity, which is acceleration.

Now if I'm given acceleration, these are derivatives right here. So what I can do is I can integrate or take the antiderivative of my acceleration to get me back to velocity. Then I can take the antiderivative of velocity to get me back to position. So that's going to be my strategy for solving this problem. So going up here to our acceleration function, notice that it's negative 9.8.

Now what I'm going to do is integrate my function right here. Now doing this antiderivative, the integral of acceleration is going to give me the velocity function with respect to time. So what I need to do is integrate this negative 9.8 here. If I integrate this negative 9.8, which is a constant with respect to t, I'm going to end up with negative 9.8t, and then don't forget the constant of integration, c. So this right here would be the general solution for my velocity.

Now what I can see here is that we're given something about the velocity. The initial velocity is 9. So we're told here that v(0) is equal to 9. So because I have this function here, I can take every place that t occurs and replace it with 0. So we're going to have that v(0) is equal to negative 9.8 times 0 plus the constant c.

Now this whole term that is being multiplied by 0 is going to go away. So all we're going to be left with is c is equal to v(0), which I can see is 9. So c is equal to 9. So that means if I rewrite this function where I take the c and replace it with 9, we're going to get that v(t) is equal to negative 9.8t plus this constant 9.

So this right here is v(t). I've gone ahead and found my velocity function with respect to time, but that's not what we're ultimately looking for in this problem. We're specifically asked to find the position function s(t). So what I can do here is recognize, well, we're going to use another initial value problem. We're going to take this s(0) here, which is equal to 0, and that's going to be our initial position.

To find the position function, I first need to find a function for position. And to do that, well, I can just integrate my velocity. The integral of velocity is going to give us s(t), the function for position with respect to time. So what I can do is integrate the right side of this equation here, and I'm going to use the rules for integrals that I know. So you can see that I'm integrating negative 9.8t, and I'm integrating 9 here.

I'm going to split this into two integrals. So we're going to have the integral of negative 9.8t with respect to t, plus and then we're going to have the integral of 9 with respect to t. Now doing this, I can also take these constants and pull them outside of the integral. So I have negative 9.8. That's going to be multiplied by the integral of t with respect to t.

You're going to have plus the integral of 9 with respect to t. Now from here, I can use the power rule on this t. Now using the power rule here, we're going to add 1 to the exponent and then divide by that new exponent. So using this power rule on the t here, we're going to end up with negative 9.8 times t^2 over 2.

Then we're going to have plus the integral of 9, which is just 9t, and then we need to add the constant of integration c. So this is what we end up getting for our function s(t). Now this is not the final answer because this is only the general solution for our position function. What we're looking for is a particular solution, a specific one. I can do that by using my initial condition up here.

So going down here, I need to take every place I see t and replace it with 0. So when we have s(0) is equal to negative 9.8 divided by 2, which, by the way, negative 9.8 divided by 2 is negative 4.9, and it's going to be multiplied by 0 squared. You're going to have plus 9 times 0 plus c. Now all these terms being multiplied by 0 are going to go away. So that means that we're just going to be left with c is equal to s(0), which we said s(0) is equal to 0, meaning that c is equal to 0.

So this c term is going to entirely go away in this function. So rewriting our position function, we're going to have s(t) is equal to this term, which we just said is negative 4.9t^2, plus, then we're going to have 9t. Then we won't have plus 0, which we don't even have to write, meaning this is our position function and the solution to this problem. So as you can see, this was a trickier problem. It was a bit more tedious.

There were more steps that we had to do. And we also had this more conceptual problem where we were given a story up here. We were told an object is moving in a line. We were given certain elements of that object, and we had to find the position. But notice how it's the same process, the same steps as we've seen with these initial value problems.

We just really need to figure out how to set it up, what our initial values are, and how we can get to our final result, which is what we did down here. So that is how you can solve these more complicated initial value problems, where there are multiple derivatives that have been taken in your function, like acceleration is the second derivative of position, and where you also have these more conceptual cases and story problems, we need to pay close attention to what's going to end up happening and what your initial conditions and values are. So, I hope you found this video helpful. Let's move on.