Finding r for 2nd Order ODE Solutions: e^rt and te^rt | Homework Help

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Homework Help Overview

The discussion revolves around finding a value of the constant r such that both e^rt and te^rt are solutions to the second-order linear homogeneous ordinary differential equation (ODE) of the form ay'' + by' + cy = 0. Participants explore the implications of repeated roots in the characteristic equation associated with the ODE.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of solutions to the ODE, questioning how e^rt and te^rt relate to the roots of the characteristic equation. There is confusion about the use of the product rule for derivatives and the necessity of substituting both forms of the solution separately.

Discussion Status

Some participants have provided hints regarding the relationship between the roots of the characteristic equation and the solutions. There is an ongoing exploration of how to verify that te^rt is indeed a solution, with various interpretations of the implications of repeated roots being discussed.

Contextual Notes

Participants express uncertainty about the steps needed to show that te^rt is a solution, particularly in relation to the derivatives and the form of the solutions. The discussion highlights the need for clarity on the conditions under which the roots of the characteristic equation are equal.

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Homework Statement


Find a value of the constant r such that both e^rt and te^rt are solutions to the ODE
ay''+by'+cy=0

Homework Equations





The Attempt at a Solution


can anyone guide me with this question please. I am 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
 
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Larrytsai said:

Homework Statement


Find a value of the constant r such that both e^rt and te^rt are solutions to the ODE
ay''+by'+cy=0

Homework Equations





The Attempt at a Solution


can anyone guide me with this question please. I am 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

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
 
so basically i sub y= e^rt + te^rt,
im confused to taking the derivative of t*e^rt,
correct me if I am wrong, i would have to use product rule on that right?
 
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.
 
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?
 
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.
 
im sorry but i don't 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:
Larrytsai said:
im sorry but i don't quite understand how to show that t * e^2t is part of the solution.
the "t" is really throwing me off.
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.
Larrytsai said:
Find a value of the constant r such that both e^rt and te^rt are solutions to the ODE
ay''+by'+cy=0
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.



Larrytsai said:
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)?

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

Homework Statement


Find a value of the constant r such that both e^rt and te^rt are solutions to the ODE
ay''+by'+cy=0

Homework Equations





The Attempt at a Solution


can anyone guide me with this question please. I am 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
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.

Larrytsai said:
so basically i sub y= e^rt + te^rt,
im confused to taking the derivative of t*e^rt,
correct me if I am wrong, i would have to use product rule on that right?
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
k so for the 2nd derivative i got

y'' = (tr^2)e^rt + 2re^rt +tre^rt + e^rt + te^rt
 
  • #11
Larrytsai said:
k so for the 2nd derivative i got

y'' = (tr^2)e^rt + 2re^rt +tre^rt + e^rt + te^rt
if i sub it into ay'' + by' + c = 0
then i get
0= t( r^2 +r +1) + 1

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
rock.freak667 said:
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.

im sorry but i don't 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
Larrytsai said:
im sorry but i don't 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.

Right, so what are the roots of ar2+br+c = 0? (Use the quadratic equation formula)
 
  • #14
[-b (+/-) sqrt ( b^2 - 4ac)]/2
 
  • #15
Larrytsai said:
[-b (+/-) sqrt ( b^2 - 4ac)]/2

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
so the determinant will make the roots equal, so they must be = 0
 
  • #17
Larrytsai said:
so the determinant will make the roots equal, so they must be = 0

Right, so that means b2-4ac=0, meaning r = ?
 
  • #18
r = -b/2a
 
  • #19
so r = -b/2a would make them both solutions.
 
  • #20
Larrytsai said:
r = -b/2a

And is this not what the question wanted you to find? :wink:
 
  • #21
okayyy i got it thnx a lot.
 

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