# Separation of Variables to Solve System Differential Equations

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Calculating Rate and Exponential Growth: The Population Dynamics Problem

### You're on a roll. Keep up the good work!

Replay
Your next lesson will play in 10 seconds
• 0:05 First Differential…
• 1:42 Second Differential…
• 6:32 Separation of Variables
• 10:32 Lesson Summary

Want to watch this again later?

Timeline
Autoplay
Autoplay

#### Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Orientation Writer
In this lesson, we discuss how to solve some types of differential equations using the separation of variables technique. We'll ponder the dastardly deeds of a mad scientist, using his chemical concoction as an example for how to use separation of variables.

## First Differential Equation Problem

Let's think for a minute about the mad scientist in some movie who's trying to take over the world. Have you ever noticed that his plans, his dastardly plans, always fail for some reason? Well often, they show some chemicals bubbling in a beaker somewhere, so I think his dastardly plans fail because he doesn't understand what chemists and chemical engineers understand, which is, he doesn't understand differential equations.

Remember that a differential equation relates a variable with its rate of change. So let's think about a chemical reaction. Now an engineer and chemist, they understand that you can write the amount of chemical that you have, let's call it A, how it changes as a function of time. So you've got dA/dt. dA/dt = -kA. So in this case, what you're doing is you're throwing chemical A into a pot, and you're letting it react. And it's going to react with a rate k times the concentration of A, where k can be any number you want. People who understand differential equations can take this differential equation, then, and determine from it the concentration of A as a function of time. So can you write A as a function of time?

## Second Differential Equation Problem

Well let's go back, and let's think about equations that we're more familiar with first. Let's say that we're looking at Super C, the human cannonball. We know that Super C starts out at a height of 0 at time t=0. When you shoot him out of the cannon, he's shot out at 13 meters per second, straight up. This means that his velocity straight up is 13 meters per second. He's always pulled down by gravity, though, which is 9.8 meters per second, squared, downwards. So given all of this information, can we determine his height as a function of time? Well first he's pulled down by gravity, which is acceleration at -9.8 meters per second squared. Acceleration is nothing more than the change in velocity over time. So I could write his acceleration at all points along his flight as being dv/dt=-9.8. It's minus because it's always being pulled back toward Earth. Now if I want to use this equation to find his actual velocity as a function of time, I'm going to multiply both sides by dt and integrate. Well, I know what the integral of dv is; it's just v. And I know what the integral of -9.8dt is; it's -9.8t. I have to add a constant of integration here, because I'm not integrating over some set time, I'm just taking an indefinite integral. So I end up with velocity equals -9.8t, plus my constant, C.

Now if I actually want to know where he is at any given point in time, I need to determine what C is. So what do I know? I know that his velocity when he was shot out of the cannon, so his velocity at time = 0, was equal to 13. So if I plug in 13 for v and 0 for t, I find that this constant, C, must be equal to 13. Okay, so we've got his velocity as a function of time. Velocity = -9.8t + 13. Can I find his position from this? Well, his velocity is dx/dt - it's how fast his position is changing with respect to time. So if I set this equal to dx/dt and multiply both sides of the equation by dt, I can integrate and find the integral of dx equals the integral of (-9.8t + 13)dt. Well, x=-4.9t^2 + 13t, plus my constant of integration. Once again, if I want to know where he is at any given point in time, I can't actually leave this constant of integration here. I need to solve for C somehow. Well, I know that at time t=0, he was at 0 height - he was just about to be shot out of the cannon, or just starting to be shot out of the cannon - so at t=0, x=0, so C has to be 0. I know that his position, his height, as a function of time, equals -4.9t^2 + 13t. So what, exactly, did I do here, other than find out how high he was at a point in time?

Well, I solved a differential equation. In particular, I solved the equation, the second derivative of x with respect to time equals -9.8, and I solved that for x as a function of time subject to what I'm going to call the initial conditions x=0 and dx/dt=13. That is, x=0, and the velocity equals 13 at time equals 0; hence, it's the initial conditions. So what does all of this have to do with determining the concentration of chemical A to avoid the dastardly fate of our dastardly mad scientist? Well, here's my differential equation: dA/dt=-kA. To solve this, to find A as a function of time, I need to integrate. But in order to integrate, I need to have t on one side of the equation and A on the other. If I just multiply both sides by dt, I still have an A on this side of the equation.

To unlock this lesson you must be a Study.com Member.

### Register for a free trial

Are you a student or a teacher?

#### See for yourself why 30 million people use Study.com

##### Become a Study.com member and start learning now.
Back
What teachers are saying about Study.com

### Earning College Credit

Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.