Differential equation: distinct, real roots

[accountkiller]

Homework Statement

Find the general solution by using either the "distinct, real roots" theorem or the "repeated roots" theorem.

9y'' - 12y' + 4y = 0

Homework Equations

"distinct, real roots" theorem - "If our roots are real & distinct, we should have solutions y₁=e^r₁x and y₂=e^r_xx, so y(x) = c₁e^r₁x + c₂^r₂x is the general solution.
"repeated roots" theorem - "If our roots are r₁=r₂, then the general solution is y(x) = (c₁+c₂x)e^r₁x.

The Attempt at a Solution

9y'' - 12y' + 4y = 0
9r² - 12r + 4 = 0
Using the quadratic formula, I got r = 2/3.
I'm confused because I only have one r. For the "distinct, real roots" theorem, the general solution is y(x) = c₁e^r₁x + c₂^r₂x, so what do I do with just my one r? Do I just leave out the second part of the general solution? Or do I use the "repeated roots" theorem since I only have one r and that general solution only includes one r?

 

[Mark44]

Homework Statement

Find the general solution by using either the "distinct, real roots" theorem or the "repeated roots" theorem.

9y'' - 12y' + 4y = 0

Homework Equations

"distinct, real roots" theorem - "If our roots are real & distinct, we should have solutions y₁=e^r₁x and y₂=e^r_xx, so y(x) = c₁e^r₁x + c₂^r₂x is the general solution.
"repeated roots" theorem - "If our roots are r₁=r₂, then the general solution is y(x) = (c₁+c₂x)e^r₁x.

The Attempt at a Solution

9y'' - 12y' + 4y = 0
9r² - 12r + 4 = 0
Using the quadratic formula, I got r = 2/3.

If you factored the equation above, you would get 9(x - 2/3)(x - 2/3) = 0. There are two roots. Are they distinct?

I'm confused because I only have one r. For the "distinct, real roots" theorem, the general solution is y(x) = c₁e^r₁x + c₂^r₂x, so what do I do with just my one r? Do I just leave out the second part of the general solution? Or do I use the "repeated roots" theorem since I only have one r and that general solution only includes one r?

 

[accountkiller]

Oh. I wasn't sure how to factor it so I just used the quadratic formula. So in order to get a correct answer, in this case the "repeated roots" theorem, I can't use the quadratic formula but have to factor?

 

[Mark44]

No, you can use the quadratic formula - you just need to recognize when you are getting repeated roots. In this case, with the characteristic equation being 9r^2 - 12r + 4 = 0, the quadratic formula gives
r = (12 +/- sqrt(144 - 144))/18, so you have r = 2/3 +/- 0. So there's a repeated root, which happens whenever the discriminant (the part in the radical) is zero.

 

[accountkiller]

Ah, I see. Thanks so much for your help! :)