Here's another example of the application of differential equations. You can imagine say someone who enjoys skydiving, jumping out of an airplane. Gravity is pulling him down, so he's under constant acceleration due to gravity, he keeps falling faster and faster. But you know that eventually his speed is going to stop increasing. That's because of the air resistance pushing up on him as he falls. So, his final velocity as free-falling skydiver is going to approach some constant velocity. He will stop accelerating. So here, I want to see if we can derive the differential equation for that, and then solve the differential equation. So, we have a skydiver, maybe it's best to draw some diagram here. So, we have some skydiver who's falling under gravity. We have to define some coordinate system. So, let's define the X-axis to be pointing up. So, the skydiver has forces on him. So, for the sake of this diagram, let's assume that his velocity is going to be down. Okay. So, the skydiver is falling. The force on the skydiver is going to be gravity. So, the gravitational force is going to be down. That's going to be minus mg, right, g is 9.8 meters per second squared. Then the air resistance, the air resistance is going to be opposing the force of gravity, so he's moving down with some velocity, so the air is pushing up. Typically, in this type of problem, you assume that the air resistance force is proportional to the velocity. So, it's a constant times is the velocity. You will always like that constant to be positive. So, here the velocity V is negative but the force is going to be up. So the force here, we can take to be minus a constant, positive constant times V. So, minus KV is positive, minus mg is negative, and then when this skydiver then is in some type of force equilibrium his velocity will be constant. So, what is the differential equation here? While we use F equals MA, right? So the Force is equal to the mass times the acceleration, and the acceleration is going to be the change in the velocity with respect to time. Then we have to set the forces. So, what are the forces here, gravity and air resistance. So, the differential equation then becomes m, dv, dt, equals the forces, gravity minus mg, and air resistance minus KV. Remember that if the diver is falling, then minus mg is negative and minus KV is positive. Okay. This is our differential equation. The independent variable is t. The dependent variable is the velocity. We see that this is a linear, first order equation, so we can put it in standard form. So, then we have dv, dt, and then we bring the plus KV over to the left side divide through by m. So, plus K over m times V, and the right-hand side is minus mg divided by m. So, It'll be minus g. That's the first-order linear equation in standard form. We have to assume some initial velocity for the skydiver. We might as well assume that he's jumping out of an airplane that's flying horizontally. So, his initial vertical velocity is zero. So, let's just take V zero, which is the vertical velocity equal to zero. Okay. So, using F equals ma, we can set up the differential equation for this problem. The only thing we need to do here is to model the effects of air resistance. So, we can analyze this equation before we actually solve it. What happens at equilibrium. So, at equilibrium the diver, the skydiver will have a balancing of forces, will stop accelerating. So, dv, dt equals zero. So, at equilibrium dv, dt equals zero. All right. So at constant velocity, equilibrium, let's call that. This dv, dt equals zero, and then V will go to what's called the terminal velocity, so that will be V infinity. That's the terminal velocity of the skydiver. You just simply put dv, dt equal to zero, solve for V, that will give us V infinity. So, V infinity will be minus mg over K. So, the terminal velocity is equal to minus mg divided by K. That will be the final velocity of the skydiver before he pulls his parachute, right? Hopefully he'll eventually pull his parachute. Okay. We can go further than this. We can ask how the skydiver approaches terminal velocity. To do that, we need to solve this differential equation, it's a linear first order equation. So, we can solve this using an integrating factor. So, we can take our integrating factor mu of t, is equal to e to the integral of P of T DT. So, it's the integral of K over M D T, which is just K T over M. So, that becomes E to the K T over M as our integrating factor. I think by this point you can do integrals of constants in your head. Then we can write down the solution for V of t. So, then V of t is equal to one over the integrating factor e to the minus KT over M, times the initial value of V which is zero plus the integral from zero to T of the integrating factor e to the K T over m times the right-hand side which is our minus g. That gives us a close form expression for the velocity. You have to do this integral, this is an easy integral to do, is just the integral of the exponential function. So, if you do this integral, and then you multiply through by e to the minus K T over M, you'll end up with minus mg over K times one minus e to the minus k t divided by m. We might as well use the terminal velocity here so the terminal velocity is minus mg over K. So, that will be the terminal velocity, which for the case of a skydiver will be negative, meaning that the sky diver's falling times one minus e to the minus K T over M. Okay. Let's interpret this expression. So, when t equals zero, the velocity is zero. The sky diver is just jumping out of the airplane, and then the skydiver is approaching the terminal velocity as this exponential decays to zero. So, it's approaching terminal velocity exponentially. In the problems you can, I ask you to work out a real example of a skydiver who's terminal velocity happens to be 200 kilometers per hour and ask you how fast he approaches terminal velocity. Okay, let me review. Sometimes you can set up a differential equation from Newton's law F equals ma. You have to consider what the forces are. In this particular problem the forces are gravity. Plus we model air resistance as being proportional to velocity. You have to be careful that you get the sign right in these terms, and then you end up with a differential equation which if we have the acceleration is d v, d t becomes a first-order equation which we can use then, this is actually separable also you can either solve this using our technique for separable equations or the fact that this is a linear first-order equation you can use our technique for linear equations. The intermediate result here is quite interesting, this is the result of equilibrium. What happens to the velocity after it stops changing when DV DT equals zero, that's the terminal velocity of this problem, and our differential equation tells us it's minus mg over K, K being related to the air resistance. Smaller air resistance faster terminal velocity. I'm Jeff Chasnov, thanks for watching and I'll see you in the next video.