If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Differential equations>Unit 2

Lesson 2: Complex and repeated roots of characteristic equation

# Repeated roots of the characteristic equations part 2

An example where we use initial conditions to solve a repeated-roots differential equation. Created by Sal Khan.

## Want to join the conversation?

• is it necessary to make C1+c2.X or c1x+c2 is also correct?
• Yes, it is correct either way. The only difference would be that c1 would be -2/3 and c2 would be 2, but they're just constants anyway, and the x is what determines their values. The more important thing is not to forget the x.
• Is reduction of order covered in this video??
• Reduction of order is covered in the previous video.
• I guess if you were given a 3rd order differential equation you would solve for the general solution the same way you would solve a cubic equation and so on for all nth order differential equations?
• Well yes and no. For the the 3rd order case you can use the cubic equation, just like Sal used the quadratic equation for this 2nd order case. But this does not extend as a strategy for the orders higher than four generally, or put otherwise there is no general solution to degree five equations, a result known as Abel-Ruffini theorem. (See link)
https://en.wikipedia.org/wiki/Abel–Ruffini_theorem
• If we set y = v*e^rx, will we always get v" = 0 so that we can get v = c1 + c2*x?
• So , if we set y=v*e^rx , and solve for r using the D.E , we find the value of r to be 1/2 and the value of y becomes v*e^x/2 and there's an error in this as the solution of a second degree equation must have two unknown constants (like c1 and c2) and that's the reason we introduce the c2*x*e^x/2 term and it also satifies the D.E , so the sum of c1*e^x/2 and c2*x*e^x/2 is the actual solution.
• at , you mentioned that because of two initial condition the solution is not general,what if we are given three initial condition? will the final solution be general?
• Well, our general solution has two arbitrary constants, so two data points are generally needed to lock down a particular solution. Now, if you tried to add a third data point, you could either end up with no solution or the same particular solution.
Think about it like this. After using the first two initial conditions, we have locked down our function. Hooray! Let's say for fun that this function is y=x^2. Now, if you gave me a third condition, say y''(0)=2, that would be consistent with our function. However, if you told me that my third condition was something like y(5)=1/2, we can no longer find the solution. y=x^2 does not pass through (1/2,5) and you will not be able to find a function that matches the differential equation as well as matches all of your initial conditions.
• You add v(x) to y(x) = v(x).e^(1/2x) because we have two initials condition what if we have three initials condition or more?
• A second order ODE only has two degrees of freedom. If you give two initial conditions, that is sufficient to establish the unique solution to the IVP. More initial conditions would be superfluous, and you might not be able to find a solution at all.
• If the original differential equation is not with respect to x, does that have any effect on the characteristic equation or for the use of roots to find the general solution?
In other words, are the methods the same, or do they vary slightly?
• You mean like if it's t instead of x? It would be exactly the same, just with t... It's just a letter.
(1 vote)
• can the C2 equals 0 in a repeated root problem?