Second order Linear ODE - NEED HELP to finish

[pat666]

Homework Statement

$$22e^{2t}=y''+8y'-9y$$

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 $$C_1e^t+C_2e^{-9t}$$
I know that r(x)=$$ke^{\gamma*x}$$ so the choice for $$y_p=Ce^{2t}$$ 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

$$22e^{2t}=y''+8y'-9y$$

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 $$C_1e^t+C_2e^{-9t}$$
I know that r(x)=$$ke^{\gamma*x}$$ so the choice for $$y_p=Ce^{2t}$$ 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.
$$y_p=Ce^{2t} y_p'=2Ce^{2t} y_p''=4Ce^{2t}$$
Then I put them back into the original ODE
so: $$4Ce^{2t}+8(2Ce^{2t})-9(Ce^{2t})$$ also the question gives two initial conditions $$y(0)=1 y'(0)=3$$ 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.
$$y_p=Ce^{2t} y_p'=2Ce^{2t} y_p''=4Ce^{2t}$$
Then I put them back into the original ODE
so: $$4Ce^{2t}+8(2Ce^{2t})-9(Ce^{2t})$$ also the question gives two initial conditions $$y(0)=1 y'(0)=3$$ 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 $$y_p=2e^(2t)$$ so $$y=C_1e^t+c_2e^-9t+2e^(2t)$$ 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, $y(t)= C_1e^t+ C_2e^{-9t}+ 2e^{2t}$ is the general solution.
But I suggest you recalculate the constants. You may have an arithmetic mistake.

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

 

[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.
$$y(0)=1$$
$$1=C_1+C_2+2$$
$$y'[t]=C_1e^t-9C_2e^{2t}+4e^{2t}$$
$$3=C_1-9C_2+4$$
$$-1+9C_2=C_1$$
$$1=-1+9C_2+c_2+2$$
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
$$y[t]=-e^t+2e^{2t}$$
Mathematicas answer is $$e^t(-1+2e^t)$$ which is exactly the same thing I think?