Second order homogenous problem

[bigu01]

Homework Statement

Solve the initial value problem y''-y'-2y=0 y(0)=β , y'(0)=2. Then find β so that the solution approaches zero as t→∞

Homework Equations

R^2-R-2=0

C1+C2=β, -C1+2C2=2

The Attempt at a Solution

I solved the equation got the r- values 1 and -2 , then i solved the two equations to find the constants, my first constant is c1=(2β-2)/3 and my second one is c2=(β+2)/3. Now I've got the equations and I don't know how to continue from here, should I try it from the first derivative (solving for minimum value or anything) or from the y-equation( it never gets to zero)

 

[tiny-tim]

hi bigu01!

hint: if you want Ce^kt -> 0 as t -> ∞, with k > 0,

then you need C = … ? ​

 

[HallsofIvy]

Homework Statement

Solve the initial value problem y''-y'-2y=0 y(0)=β , y'(0)=2. Then find β so that the solution approaches zero as t→∞

Homework Equations

R^2-R-2=0

C1+C2=β, -C1+2C2=2

The Attempt at a Solution

I solved the equation got the r- values 1 and -2

Well, here's your first problem. r= 1 and r= -2 do NOT satisfy the equation:
(1)^2- 1- 2= -2, not 0, and (-2)^2- (-2)- 2= 4, not 0.

, then i solved the two equations to find the constants, my first constant is c1=(2β-2)/3 and my second one is c2=(β+2)/3. Now I've got the equations and I don't know how to continue from here, should I try it from the first derivative (solving for minimum value or anything) or from the y-equation( it never gets to zero)

 

[bigu01]

Well, here's your first problem. r= 1 and r= -2 do NOT satisfy the equation:
(1)^2- 1- 2= -2, not 0, and (-2)^2- (-2)- 2= 4, not 0.

The R-s are inputed to the e^rt not directly in the equation right, so if you input the r=1 and r=-2 ,
y becomes y=c1e^t+c2e^-2t

 

[Mark44]

The R-s are inputed to the e^rt not directly in the equation right, so if you input the r=1 and r=-2 ,
y becomes y=c1e^t+c2e^-2t

No.

Your characteristic equation is r² - r - 2 = 0. The solutions are NOT r = 1 and r = -2. This is what HallsOfIvy is saying.

Try again.

 

[bigu01]

No.

Your characteristic equation is r² - r - 2 = 0. The solutions are NOT r = 1 and r = -2. This is what HallsOfIvy is saying.

Try again.

Oh I see r=-1 and r=2 I solved it like that in my notebook,but when putting here looks like I've mixed them.Sorry.Then I have (β+2)/3*e^2t+(2β+2)*e^-t =y and I have to get y=0 as t->∞

 

[bigu01]

hi bigu01!

hint: if you want Ce^kt -> 0 as t -> ∞, with k > 0,

then you need C = … ? ​

k=e^-t?

 

[tiny-tim]

sorry, you're not making any sense

 

[bigu01]

sorry, you're not making any sense

β=-2 C should be 0 , β+2=0

 

[tiny-tim]

C should be 0 , β+2=0

that's better!