# 2nd order ODE

## 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

## 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

rock.freak667
Homework Helper

## 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

## 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

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?

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?

Mark44
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.

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:
Mark44
Mentor
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).

HallsofIvy
Homework Helper

## 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

## 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.

k so for the 2nd derivative i got

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

rock.freak667
Homework Helper
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.

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.

rock.freak667
Homework Helper
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)

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

rock.freak667
Homework Helper
[-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).

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

rock.freak667
Homework Helper
so the determinant will make the roots equal, so they must be = 0

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

r = -b/2a

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

rock.freak667
Homework Helper
r = -b/2a

And is this not what the question wanted you to find? okayyy i got it thnx alot.