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### 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?
• You look at your equation, C1 is the first C and C2 is the second. It doesn't actually doesn't matter if you labeled the first one as C2 or the second one as C8 as long as you continue using that placement of first and second once you start solving for C1,2,3....n
• 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?
• Adding or subtracting those equations depends on the sign of c1 or c2 . Our aim is to cancel out one of the constants (either c1 or c2) and thereby find the other. Here we have the sign of c1 positive in both the equations.c1 cancels out only when we subtract. hope you understand
(1 vote)
• 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)
• Just add 1 to both sides so it becomes 5y''+3y'+4y=1 then apply undetermined coefficients or laplace transforms to compute the answer.