# 2nd order ODE

1. Sep 25, 2010

### Larrytsai

1. The problem statement, all variables and given/known data
Find a value of the constant r such that both e^rt and te^rt are solutions to the ODE
ay''+by'+cy=0

2. Relevant equations

3. The attempt at a solution
can anyone guide me with this question please. Im not sure where to start.
I know that e^rt is always a solution with homogeneous eqns, so in order for te^rt to be
a solution it must equal e^rt
so...

e^rt = te^rt

2. Sep 25, 2010

### rock.freak667

If your solutions are y1=ert and y2=tert, then what does that say about the roots of the auxiliary equation?

Hint: Look at the form of the solution of y=Ay1+By2

3. Sep 25, 2010

### Larrytsai

so basically i sub y= e^rt + te^rt,
im confused to taking the derivative of t*e^rt,
correct me if im wrong, i would have to use product rule on that right?

4. Sep 25, 2010

### JThompson

If you have a second order differential equation with constant coefficients, such as ay''+by'+cy=0, then you can find r in the solutions you gave by finding the roots of the characteristic equation ar^2+br+c=0. The solutions you presented are characteristic of a repeated root. A "repeated root" means that the determinant in the quadratic equation is zero.

5. Sep 25, 2010

### Larrytsai

the roots for e^rt is just [-b(+/-)sqrt(b^2-4ac)]/2a
how would i be able to find the equivalent for t * e^rt?

6. Sep 25, 2010

### Staff: Mentor

If you have repeated roots, then the discriminant (b^2 - 4ac) in the quadratic formula will be zero.

For example, the quadratic equation x^2 - 4x + 4 = 0 has x = 2 as a repeated root.

If the differential equation were y'' - 4y' + 4y = 0, the characteristic equation would be r^2 - 4r + 4 = 0, which has repeated roots.

For this differential equation, we need two linearly independent functions, and these will be y1 = e^(2t) and y2 = te(2t). You can easily check that both are solutions to y'' - 4y' + 4y = 0.

7. Sep 25, 2010

### Larrytsai

im sorry but i dont quite understand how to show that t * e^2t is part of the solution.
the "t" is really throwing me off.
So i was wondering if i want those 2 solutions to be the same, the 'r' must have the same roots, so...

r=-b/2a
then
y1 = e^(-bt/2a)
and
y2 = t*e^(-bt/2a)?

Last edited: Sep 25, 2010
8. Sep 25, 2010

### Staff: Mentor

How do you usually verify that a solution to a differential equation is actually a solution? Plug it into the diff. equation and see if you get a true statement, that's how.

For the DE that I gave as an example, y'' -4y' + 4y = 0, check that y = te2t is a solution.

y = te2t
y' = e2t + 2te2t
y'' = 2e2t + 2e2t + 4te2t = 4e2t + 4te2t

Then y'' - 4y' + 4y = (4e2t + 4te2t) - 4(e2t + 2te2t) + 4te2t
= 0. This is true for all values of t.

Therefore, y = te2t is a solution to the differential equation y'' - 4y' + 4y = 0.

I think that you are losing sight of what you need to do in this problem.
You have found a value of r so that you get two identical (repeated) roots in the characteristic equation (ar2 + br + c = 0).

Now all you need to do is to show that y = tert is a solution to the diff. equation ay'' + by' + cy = 0.

These two solutions are different, not the same, but they use the same value of r, which is -b/(2a).

9. Sep 26, 2010

### HallsofIvy

No, that does not follow. For one thing you do NOT know that any solution must be of the form "e^{rt}" and no one has told you that! It was probably recommended in your text that you try something of the form e^{rt} (for a linear homogenous equation with constant coefficents) and then nice things happen- your differential equation reduces to a polynomial equation- the "characteristic equation". But it was never said that the the solution must of that form. You get different kinds of solutions in the case of a multiple (or triple, etc) root or complex roots to the characteristic equation.

Yes, of course. that is a product of t and e^rt. Its first derivative is e^rt+ rte^rt. Can you find the second derivative?

However, you don't need to substitute y= e^rt+ te^rt. Instead look at y= e^rt and y= te^rt separately. IF they are both solutions, then the general solution is a linear combination of them.

10. Sep 26, 2010

### Larrytsai

k so for the 2nd derivative i got

y'' = (tr^2)e^rt + 2re^rt +tre^rt + e^rt + te^rt

11. Sep 26, 2010

### rock.freak667

You did not even need to do all of that, subbing y=ert would give you the auxiliary equation, since the roots are ert and tert, it means the roots of the auxiliary equation are equal.

12. Sep 26, 2010

### Larrytsai

im sorry but i dont quite understand, i subbed e^rt in as u said and got ar^2 + br + c = 0
then the roots would be just the quadratic equation.

13. Sep 26, 2010

### rock.freak667

Right, so what are the roots of ar2+br+c = 0? (Use the quadratic equation formula)

14. Sep 26, 2010

### Larrytsai

[-b (+/-) sqrt ( b^2 - 4ac)]/2

15. Sep 26, 2010

### rock.freak667

Now your solutions are y=ert and y=tert, meaning that the roots are real and equal.

In r= [-b± √(b2 - 4ac)]/2a, what condition would make the roots equal? (Look what is under the square roots sign).

16. Sep 26, 2010

### Larrytsai

so the determinant will make the roots equal, so they must be = 0

17. Sep 26, 2010

### rock.freak667

Right, so that means b2-4ac=0, meaning r = ?

18. Sep 26, 2010

### Larrytsai

r = -b/2a

19. Sep 26, 2010

### Larrytsai

so r = -b/2a would make them both solutions.

20. Sep 26, 2010

### rock.freak667

And is this not what the question wanted you to find?