2nd order ODE

  • Thread starter Larrytsai
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  • #1
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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. 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
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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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. 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

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
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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
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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
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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
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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
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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:
  • #8
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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.
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.



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).
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,847
966

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. 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
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.

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?
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
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k so for the 2nd derivative i got

y'' = (tr^2)e^rt + 2re^rt +tre^rt + e^rt + te^rt
 
  • #11
rock.freak667
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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
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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 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
rock.freak667
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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.

Right, so what are the roots of ar2+br+c = 0? (Use the quadratic equation formula)
 
  • #14
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[-b (+/-) sqrt ( b^2 - 4ac)]/2
 
  • #15
rock.freak667
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[-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
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so the determinant will make the roots equal, so they must be = 0
 
  • #17
rock.freak667
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so the determinant will make the roots equal, so they must be = 0

Right, so that means b2-4ac=0, meaning r = ?
 
  • #18
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r = -b/2a
 
  • #19
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so r = -b/2a would make them both solutions.
 
  • #20
rock.freak667
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r = -b/2a

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

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