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### Course: Differential equations>Unit 2

Lesson 2: Complex and repeated roots of characteristic equation

# Repeated roots of the characteristic equation

What happens when the characteristic equation only has 1 repeated root? Created by Sal Khan.

## Want to join the conversation?

• This video doesn't seem to work anymore?
• I had the same problem and this advice might help, If you drag the round button (above play) and forward it to n seconds the video might start working. I have a google chrome browser.
• In the first video on 2nd order DE Sal gave us general solution for them and told that this was the only solution and there is no other. I took it as a leap of faith. I don't like it, and it's pretty much why I like mathemathics. Now we are told that actually there is case when that solution is no good. That is so confusing. Could anyone enlighten me why that previous solution was fine there and is not fine anymore here? I don't want to just learn solutions to equations by heart, I want to understand it all and at the moment those 2nd order DEs are super confusing.
• Around the mark, dont know how to make the quick link here, he says 2nd order Homogenous equations must have two solutions for the general solution. Later on it is shown they must be unique. If not, a contradiction is presented around . 2nd order means two unique solutions for the general solution and two initial conditions to find the C's. Hope that helps a bit.
• at Sal says we need to differentiate twice to get the solution, but he meant that we need to take the anti-derivative twice.
• How do we know that the solution v(x)*exp(-2x) is the most general possible?
• consider if v(x) contained the factor exp(2x). Then it would cancel with the exp(-2x). No solutions are actually neglected by making this assumption.

The assumption that I am still trying hard not to simply "come to terms with" is the assumption for 2nd order equations that solutions will be of the form y=exp(kx) in the first place. Is there a video that addresses this?
• Saying that y=ce^(rx) isn't the solution because you can pick two initial conditions which this solution can't satisfy isn't a very convincing argument. I know that Sal's only bringing it up to give you some intuition, not as a proof, but I would still like something a little more concrete.

For any second order linear homogeneous equation I could pick three initial conditions that the general solution could not simultaneously satisfy, one for y, one for y', and one for y''. So just saying that you could pick some initial conditions which this general solution couldn't satisfy isn't very... satisfying.
• why do we used some function V(x) multiplied by the first solution why not adding a constant or adding the function ?
• Why did Sal take the general solution to be a product of v(x) and e^-2x ? Why not their sum ?
• In the beginning of the video Sal tried to just use a constant which we saw didn't work. By changing the constant into a 'constant function', v(x), you can use the chain rule and figure out the general solution. So instead of multiplying e^-2x by a constant, he multiplied by v(x), a 'constant function'
• When sal was integrating last part V(X) why didnt he change the sign??

I mean V''(x)=0,

V'(x) + c1 =0

V(x) + c1*x + c2 =0

so V(x) = -c1*x -c2

Is that because this is just constant and the sign dont care about?
• It's true that the constants can be chosen arbitrarily, so they account for all the negative solutions as well and so there's no need for any negative signs out front. What confuses me though is why you introduce the negatives in the first place by placing the constants from integration on the left side as opposed to the right.
• At the expression simplifies into v'' = 0, a separable second order differential equation. Why is this technique called "reduction of order", if this new differential equation is still second order?
• In the full "Reduction of order" procedure, once you have substituted your possible general solution into the original equation, you have to then make a variable substitution `w = v'`, since the method guarantees that the 0-order derivative terms will have vanished. After that substitution you end up with a first order differential equation.

In this case, since both the 0-order and the 1-order differential terms disappeared, there was no need to make the substitution to continue working.

You can see the full method here: http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx