It should work. Eg. C = -1/3. Post your working.
Note: The same DE was posted by a different member at this thread: http://www.mathhelpforum.com/math-he...equations.html
hi, i am havin alot of trouble gettin the general solution to the ODE:
y'' - 7y' + 6y = 2(e^(3x)) + x(e^x)
i tried using the function for the particular solution as : C(e^3x) + (Ax+B)e^x but it doesnt work, like some of the constants cancel out and i cant get a proper solution!
how would i go about doin this question?
It should work. Eg. C = -1/3. Post your working.
Note: The same DE was posted by a different member at this thread: http://www.mathhelpforum.com/math-he...equations.html
yes i got c = -1/3 and also got A = -t/5 but i cant get a value for B.
i ended up with -5Ae^t = te^t
i had another expression (At+B)e^t but they all canceled out.
so now i have for y = (At+B)e^t = ((-t^2)/5 + B)e^t
what happens to the B? does it jus equal zero or something? im not sure?
ok sorry. i did mean A = -t/5 sorry i ended up using t's instead of x's coz of that post u linked me to.
well for y'' - 7y' + 6y = t(e^t) (i have taken out 2(e^(3t)) since i have no problem with that) - now im jus goin with t's
what i did was set y = (Ax+B)e^t
and then y' = (Ax+B)e^t + Ae^t
and then y'' = (Ax+B)e^t + 2Ae^t
so we have for y'' -7y' +6y :
(Ax+B)e^t + 2Ae^t - 7[(Ax+B)e^t + Ae^t] + 6[(Ax+B)e^t]
which when expanded is:
(Ax+B)e^t + 2Ae^t - 7(Ax+B)e^t - 7Ae^t + 6(Ax+B)e^t
which simplifies to:
-5Ae^t
so i equated the term to t(e^t) to get:
-5Ae^t = te^t
-5A = t
A = -t/5
and now i put that into y = (At+B)e^t = ((-t^2)/5 + B)e^t
so question is, what do i do with the B?
also i am sorry for writing it out with t's when i wrote the initial question in x's....
My mistake. Clearly y = (At + B) e^t = At e^t + B e^t is redundant since B e^t is part of the homogenous solution.
So you try y = A t e^t. This doesn't work because you end up with -5 A e^t on the left and t e^t on the right. By definition A is a constant and so things aren't right.
So try y = (A t^2 + Bt) e^t as a particular solution .....
ahh ive tried y = (A t^2 + Bt) e^t as well and i jus get it down to:
2A-10At - 5B = t
I also tried y = (At^2 + Bt + C) e^t but get the same result as above. I tried this one too because from the answer galactus got there seems to be 3 terms for e^t.
I dont think iam doing it right (i did it same method as i showed u previously)...could someone show me how it is done using undetermined coefficients, i would really appreciate it coz im pretty lost now lol
ok wait sorry i jus realised what i did.
i had the answer:
from 2A-10At - 5B = t
we equate -10At = t to get A = -1/10
and then 2A - 5B = 0
and then B = -1/25
so now we have for y= (At^2 + Bt)e^t = (-1/10t^2 -1/25t)e^t
so it works BUT why is my answer different to galactus???
i dont end up with a (-1/125e^t) term??