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## Differential equations

### Course: Differential equations>Unit 2

Lesson 1: Linear homogeneous equations

# 2nd order linear homogeneous differential equations 4

Another example with initial conditions! Created by Sal Khan.

## Want to join the conversation?

• Now that I've been through this playlist, I have to ask, will the general solutions for all linear homogeneous differential equations include e^rx? I mean, are there usually no other types of solutions that appear in these problems (like maybe some arbitrary polynomial or perhaps even natural log), just the exponential function? Thanks in advance. •  In short, no. A polynomial kind of degrades as you take successive derivatives, so the derivatives cannot sum to 0. Differentiating natural log puts it in the form 1/x, so you definitely could not get things to cancel out.
However, there are non-homogeneous ODEs. These are of the form
Ay''+By'+Cy=f(x). These equations end up having a solution that can be a combination of the exponential function and polynomials, trigonometric, etc. functions. Check out the videos on the method of undetermined coefficients. They're really interesting!
• I tried doing some research online, and I kept on reading about something called the Wronskian. Could you please explain that? • Right, so if we have n solutions to a differential equation that are each differentiable n-1 times, the Wronskian is the determinant of the matrix formed by the equations and each of their k derivatives. So, if we have two solutions, y_1 and y_2 (n =2), and they are differentiable so that we can get y_1' and y_2' then we can form the Wronskian as follows:
| y_1 y_2 |
W = | y_1' y_2' | = y_1*y_2' - y_1'*y_2

This is the simple case of n=2, and you can show that it expands to be true for all n if you assume the equations that form the solution to be 'nice enough'.
• What if a Constant (C1 or C2) ends up equaling 0... which e (and it's power) am suppose to eliminate? • Do you add og subtract when you do the thing with:
c1+c2=2
3/2c1+1/2c2=1/2
On this vid you subtract, and on the video before you add them together? • Using the same technique, can you solve first order differential equations in the same fashion, 0y'+3y'+6y=0? Can you solve a first order equation that way? • Short answer, yes. In fact any constant coefficient equation like the one you gave, the one in this video, or one of any order is really only as hard as finding the roots of its characteristic equation. These get harder as their degree goes up, but you could cook up some easy third-order examples. Their characteristic equations would have a degree matching the order, so would be cubics. Hope this helps!
• You can solve the quadratic equation without using the quadratic formula like this:
4r^2 - 8t + 3 = 0 |Divide by four:
r^2 - (8/4)r + 3/4 = 0 |Multiply both numerator and denominator of 3/4 by four:
r^2 - (8/4)r + 12/16 = 0 |Figure out two numbers a,b that satisfy a + b = -8 and ab = 12:
a = -2, b = -6 |Substitute a,b into (r + a/4) (r + b/4) = 0:
(r - 2/4) (r - 6/4) = 0
(r - 1/2) (r - 3/2) = 0
r = 1/2 or r = 3/2 • Are there any videos on this site about systems of differential equations? • What if the values for r are the same? How do you solve for C1 and C2 then?

Example:

y(0) = C1 + C2 = 5
y'(0) = (-13/5)C1 + (-13/5)C2 = 5

How would you go about solving this one? • Interesting question!

The system of equations that you wrote has no solutions, because C1 + C2 = 5 would imply that (-13/5)C1 + (-13/5)C2 equals -13 instead of 5. So there's a problem.

I suspect from your system of equations that you got a double root of r = -13/5 when solving a second order differential linear homogeneous equation, and the initial conditions were y(0) = 5, y'(0) = 5.
In this situation, the general solution is y = C1 e^(-13x/5) + C2 x e^(-13x/5), because one of the two terms of the general solution needs a factor of x when there's a double root.

Then the initial conditions y(0) = 5 and y'(0) = 5 would lead to the system
C1 = 5
(-13/5)C1 + C2 = 5.
This system of equations is quite easy to solve.

Have a blessed, wonderful day!
(1 vote)
• Hi, understand this. Thank you very much for the great explanation. I have a question thought:
can someone tell me what I would do with an equation in this form:
5y'' + 3y' +4y - 1 = 0 ?
What to do with the "-1" ? Can I neglect it? Am i missing an obvious algebraic transformation?
(1 vote) • At is multiplying only top equation right? It seems intuitive, but doesn't that affect the solution?
(1 vote) ## Video transcript

Let's solve another 2nd order linear homogeneous differential equation. And this one-- well, I won't give you the details before I actually write it down. So the differential equation is 4 times the 2nd derivative of y with respect to x, minus 8 times the 1st derivative, plus 3 times the function times y, is equal to 0. And we have our initial conditions y of 0 is equal to 2. And we have y prime of 0 is equal to 1/2. Now I could go into the whole thing y is equal to e to the rx is a solution, substitute it in, then factor out e to the rx, and have the characteristic equation. And if you want to see all of that over again, you might want to watch the previous video, just to see where that characteristic equation comes from. But in this video, I'm just going to show you, literally, how quickly you can do these type of problems mechanically. So if this is our original differential equation, the characteristic equation is going to be-- and I'll do this in a different color-- 4r squared minus 8r plus 3r is equal to 0. And watch the previous video if you don't know where this characteristic equation comes from. But if you want to do these problems really quick, you just substitute the 2nd derivatives with an r squared, the first derivatives with an r, and then the function with-- oh sorry, no. This is supposed to be a constant-- And then the coefficient on the original function is just a constant, right? I think you see what I did. 2nd derivative r squared. 1st derivative r. No derivative-- you could say that's r to the 0, or just 1. But this is our characteristic equation. And now we can just figure out its roots. This is not a trivial one for me to factor so, if it's not trivial, you can just use the quadratic equation. So we could say the solution of this is r is equal to negative b-- b is negative 8, so it's positive 8-- 8 plus or minus the square root of b squared. So that's 64, minus 4 times a which is 4, times c which is 3. All of that over 2a. 2 times 4 is 8. That equals 8 plus or minus square root of 64 minus-- what's 16 times 3-- minus 48. All of that over 8. What's 64 minus 48? Let's see, it's 16, right? Right. 10 plus 48 is 58, then another-- so it's 16. So we have r is equal to 8 plus or minus the square root of 16, over 8, is equal to 8 plus or minus 4 over 8. That equals 1 plus or minus 1/2. So the two solutions of this characteristic equation-- ignore that, let me scratch that out in black so you know that's not like a 30 or something-- the two solutions of this characteristic equation are r is equal to-- well 1 plus 1/2 is equal to 3/2-- and r is equal to 1 minus 1/2, is equal to 1/2. So we know our two r's, and we know that, from previous experience in the last video, that y is equal to c times e to the rx is a solution. So the general solution of this differential equation is y is equal to c1 times e-- let's use our first r-- e to the 3/2 x, plus c2 times e to the 1/2 x. This differential equations problem was literally just a problem in using the quadratic equation. And once you figure out the r's you have your general solution. And now we just have to use our initial conditions. So to know the initial conditions, we need to know y of x, and we need to know y prime of x. Let's just do that right now. So what's y prime? y prime of our general solution is equal to 3/2 times c1 e to the 3/2 x, plus-- derivative of the inside-- 1/2 times c2 e to the 1/2 x. And now let's use our actual initial conditions. I don't want to lose them-- let me rewrite them down here so I can scroll down. So we know that y of 0 is equal to 2, and y prime of 0 is equal to 1/2. Those are our initial conditions. So let's use that information. So y of 0-- what happens when you substitute x is equal to 0 here.? You get c1 times e to the 0, essentially, so that's just 1, plus c2-- well that's just e to the 0 again, because x is 0-- is equal to-- so this is, when x is equal to 0, what is y? y is equal to 2. Y of 0 is equal to 2. And then let's use the second equation. So when we substitute x is equal to 0 in the derivative-- so when x is 0 we get 3/2 c1-- this goes to 1 again-- plus 1/2 c2-- this is 1 again, e to the 1/2 half times 0 is e to the 0, which is 1-- is equal to-- so when x is 0 for the derivative, y is equal to 1/2, or the derivative is 1/2 at that point, or the slope is 1/2 at that point. And now we have two equations and two unknowns, and we could solve it a ton of ways. I think you know how to solve them. Let's multiply the top equation-- I don't know-- let's multiply it by 3/2, and what do we get? We get-- I'll do it in a different color-- we get 3/2 c1 plus 3/2 c2 is equal to-- what's 3/2 times 2? It's equal to 3. And now, let's subtract-- well, I don't want to confuse you, so let's just subtract the bottom from the top, so this cancels out. What's 1/2 minus 3/2? 1/2 minus 1 and 1/2. Well, that's just minus 1, right? So minus c2 is equal to-- what's 1/2 minus 3? It's minus 2 and 1/2, or minus 5/2. And so we get c2 is equal to 5/2. And we can substitute back in this top equation. c1 plus 5/2 is equal to 2, or c1 is equal to 2, which is the same thing as 4/2, minus 5/2, which is equal to minus 1/2. And now we can just substitute c1 and c2 back into our general solution and we have found the particular solution of this differential equation, which is y is equal to c1-- c1 is minus 1/2-- minus 1/2 e to the 3/2 x plus c2-- c2 is 5/2-- plus c2, which is 5/2, e to the 1/2 x, and we are done. And it might seem really fancy. We're solving a differential equation. Our solution has e in it. We're taking derivatives and we're doing all sorts of things. But really the meat of this problem was solving a quadratic, which was our characteristic equation. And watch the previous video just to see why this characteristic equation works. But it's very easy to come up with the characteristic equation, right? I think you obviously see that y prime turns into r squared, y prime turns into r, and then y just turns into 1, essentially. So you solve a quadratic. And then after doing that, you just have to take one derivative-- because after solving the quadratic, you immediately have the general solution-- then you take its derivative, use your initial conditions. You have a system of linear equations which is Algebra I. And then you solve them for the two constants, c1 and c2, and you end up with your particular solution. And that's all there is to it. I will see you in the next video.