Second order Linear ODE - NEED HELP to finish

[pat666]

Homework Statement

[tex]
22e^{2t}=y''+8y'-9y
[/tex]

Homework Equations

The Attempt at a Solution

The directly previous question to this was the same but homogeneous, i.e. the 22e^(2t) was replaced with a 0.
So I know the general solution to the homogeneous ode is [tex] C_1e^t+C_2e^{-9t} [/tex]
I know that r(x)=[tex] ke^{\gamma*x} [/tex] so the choice for [tex]y_p=Ce^{2t} [/tex] this next part is what I don't understand. I think I just use the Basic rule but an example very similar to mine uses the modification rule??

Please help

Thanks

 

[LCKurtz]

Homework Statement

[tex]
22e^{2t}=y''+8y'-9y
[/tex]

Homework Equations

The Attempt at a Solution

The directly previous question to this was the same but homogeneous, i.e. the 22e^(2t) was replaced with a 0.
So I know the general solution to the homogeneous ode is [tex] C_1e^t+C_2e^{-9t} [/tex]
I know that r(x)=[tex] ke^{\gamma*x} [/tex] so the choice for [tex]y_p=Ce^{2t} [/tex] this next part is what I don't understand. I think I just use the Basic rule but an example very similar to mine uses the modification rule??

Please help

Thanks

Since derivatives of e^2t always have an e^2t in their answer, it seems logical that if you want to get a constant times e^2t out on the right (usually, in your case, on the left) you should try a constant times e^2t for y to make the equation work. The only time this wouldn't work is if e^2t is a solution to the homogeneous equation because then you would get 0. That is when you modify your guess to te^2t.

 

[pat666]

Hey,
By constant do you mean 3e^2t or Ce^2t?

thanks

 

[LCKurtz]

Hey,
By constant do you mean 3e^2t or Ce^2t?

thanks

You try
Ce^2t and figure out what C makes the equation work.

 

[pat666]

Yep that's what I thought you meant, I tried that before posting and it didn;t work for me at all.
[tex] y_p=Ce^{2t} y_p'=2Ce^{2t} y_p''=4Ce^{2t} [/tex]
Then I put them back into the original ODE
so: [tex] 4Ce^{2t}+8(2Ce^{2t})-9(Ce^{2t}) [/tex] also the question gives two initial conditions [tex] y(0)=1 y'(0)=3 [/tex] could you tell me what I've done wrong??

Thanks

 

[LCKurtz]

Yep that's what I thought you meant, I tried that before posting and it didn;t work for me at all.
[tex] y_p=Ce^{2t} y_p'=2Ce^{2t} y_p''=4Ce^{2t} [/tex]
Then I put them back into the original ODE
so: [tex] 4Ce^{2t}+8(2Ce^{2t})-9(Ce^{2t}) [/tex] also the question gives two initial conditions [tex] y(0)=1 y'(0)=3 [/tex] could you tell me what I've done wrong??

Thanks

You put them back both sides of the equation. The other side of the equation is 22e^2t. Once you have figured out what C works you know y_p = Ce^2t with that value of C. Then your general solution is y = the general solution to your homogeneous equation + y_p. Only then do you go to work on the initial conditons.

 

[pat666]

Equating both sides such that I have 11Ce^(2t)=22e^(2t) its pretty obvious that C = 2.
so now I have [tex]y_p=2e^(2t)[/tex] so [tex]y=C_1e^t+c_2e^-9t+2e^(2t) [/tex] then I solve for C1 and C2 and get C1=-0.8 and C2=-0.2, I know this is wrong because Mathematica solved it as y[t]=e^t(-1+2e^(2t)) what have I stuffed up?

Thanks

 

[HallsofIvy]

Yes, [itex]y(t)= C_1e^t+ C_2e^{-9t}+ 2e^{2t}[/itex] is the general solution.
But I suggest you recalculate the constants. You may have an arithmetic mistake.

[itex]y'(t)= C_1 e^t- 9C_2e^{-9t}+ 4e^2t[/itex] so we must have
[itex]y(0)= C_1+ C_2+ 2= 1[/itex] and [itex]y'(0)= C_1- 9C_2+ 4= 3[/itex].

 

[pat666]

I have recalculated several times, I can't figure it out. I know I'm wrong because I have done it in mathematica. Ill write out what I did here so you might be able to tell where I've gone wrong.
[tex]y(0)=1[/tex]
[tex]1=C_1+C_2+2 [/tex]
[tex]y'[t]=C_1e^t-9C_2e^{2t}+4e^{2t}[/tex]
[tex]3=C_1-9C_2+4[/tex]
[tex]-1+9C_2=C_1[/tex]
[tex]1=-1+9C_2+c_2+2[/tex]
Now I keep getting C_2=0 which is wrong!Thanks

 

[pat666]

Whoops, yes C_2 should = 0 and C_1=-1.

Problem solved I think
[tex]y[t]=-e^t+2e^{2t}[/tex]
Mathematicas answer is [tex]e^t(-1+2e^t) [/tex] which is exactly the same thing I think?