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

[Larrytsai]

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

 

[rock.freak667]

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=e^rt and y₂=te^rt, then what does that say about the roots of the auxiliary equation?

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

 

[Larrytsai]

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?

 

[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.

 

[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?

 

[Mark44]

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.

 

[Larrytsai]

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)?

 

[Mark44]

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 = te^2t is a solution.

y = te^2t
y' = e^2t + 2te^2t
y'' = 2e^2t + 2e^2t + 4te^2t = 4e^2t + 4te^2t

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

Therefore, y = te^2t 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.

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 = te^rt 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 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.

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.

 

[Larrytsai]

k so for the 2nd derivative i got

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

 

[rock.freak667]

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=e^rt would give you the auxiliary equation, since the roots are e^rt and te^rt, it means the roots of the auxiliary equation are equal.

 

[Larrytsai]

You did not even need to do all of that, subbing y=e^rt would give you the auxiliary equation, since the roots are e^rt and te^rt, 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.

 

[rock.freak667]

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)

 

[Larrytsai]

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

 

[rock.freak667]

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

Now your solutions are y=e^rt and y=te^rt, 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).

 

[Larrytsai]

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

 

[rock.freak667]

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

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

 

[Larrytsai]

r = -b/2a

 

[Larrytsai]

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

 

[rock.freak667]

r = -b/2a

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

 

[Larrytsai]

okayyy i got it thnx alot.