Question

Solve the differential equation
y''+2y'+2y=sin(t)e^(-t)

I'm mainly having problems with the practical solution and wondering if there is a better way of doing it. The practical solution i'm thinking is

y(prac)=t*e^(-t)*(A*cos(t)+B*sin(t))

Will i need to take derivatives and put back into the original equation then solve or is there a better method for doing this ?

Answer 1

I think \(y_p=e^{-t}(A\cos t+B\sin t)\) would work.

Answer 2

solve your characteristic equation, good so your solution will be \[y=-1/2 e^{-t}tcos(t) c1e^{-t}\cos(t)+c2e^{-t}\sin(t) \]

Answer 3

m^2+2m+2 = -1+/-i

Answer 4

Or not. It's not right !

Answer 5

@Anwara it has to have t as well as it is the same as the complimentary equation otherwise. I'm stuck solving the (Acos(t) + Bsin(t)) in the y(prac). I'm wondering if i need to take derivatives of the y(prac) and then put this back into my original equation and solve for A and B or if there is a better way. As the derivatives are rather long

Answer 6

@JohnnyBreur sorry i should have been more clear. The issue i'm having is solving the A and B in the y(p) must i take the derivatives and put them back into the original equation and solve for A and B or is there a better method

Answer 7

back sub to solve for A and B

Answer 8

you start with y''+2y'+2y=0n find your solution

Answer 9

Therefore i have to solve for y'', y' of the prac to back sub and find my solution. e.g.
\[y_p'=\exp(-t)(-A t \sin(t)+A \cos(t)-A t \cos(t)+B \sin(t)-B t \sin(t)+B t \cos(t))\]

And solve y'' and then insert into the equation? I was just trying to find a method that wouldnt involve such a large equation if there were one otherwise i'll just have to do it this way

Answer 10

Do you know the annihilator approach? It might be easier to use it in this case.

Answer 11

No i'm unfamilar with the annihilator approach did you have a good link to info on it? Might be worth a look

Answer 12

http://math.wallawalla.edu/~duncjo/courses/math312/spring05/notes/ode_chapter_4-5b.pdf

Answer 13

i think you need an annihilator here
(D^2-2aD+a^2+b^2)
this will annihilate e^atsin(bt)
since a=(-1)
b=1
(D^2-2(-1)D+1+1)
sin(t)e^(-t)
D=cos(t)e^-t-sin(t)e^-t
D^2=-sin(t)e^-t-cos(t)e^-t-cos(t)e^-t+sin(t)e^-t
2D=(2cos(t)e^-t-2sin(t)e^-t)
D^2=(-2cos(t)e^-t
THE annihilator does not look like its going to 0
(-2cos(t)e^-t+2)
you think i went wrong somewhere ANWARA?