Question

diff eq..determine the form of a particular solution for the given nonhomogeneous equation
y''+2y'=3-4t

Answer 1

yp = some educated guess

yp=At + B ; if these are not already a linear solution to the homogenous part ...

Answer 2

for fundamental solution set, i have y1=1, y2=e^(-2t)
f(t)=3-4t
polynomial method so
yp=-At+B...check to see if any matches the fund.set. it does so...
yp=-At^2+Bt
find y'p and y''p and plug in. now solving for A and B is where i'm stuck

Answer 3

then lets work thru it so I can catch up :)

Answer 4

ok thx

Answer 5

r^2+2r=0

r=0,-2 right?

Answer 6

at this point, it's just algebra which apparently, i'm very weak at

Answer 7

right

Answer 8

\[y=c_1e^{0x}+c_2e^{-2x}+y_p\]
\[y=c_1+c_2e^{-2x}+y_p\]

Answer 9

yp = At + B ; but we already have a constant so we need to up the ante

yp = At^2 + Bt ; and derive

Answer 10

yp = At^2 + Bt
yp'= 2At + B
yp''= 2A

agreed?

Answer 11

yp''+2yp'=3-4t

(2A) + 2(2At + B)=3-4t

im going to do this in lowercase, just easier to type for me

2a +4at +2b = 3 -4t

(4a)t = -4t ; a=-1

(2a+2b) = 3

-2+2b = 3 ; b=5/2

Answer 12

ah ok, i almost did that to solve for a but wasn't sure if it was correct, thank you

Answer 13

at^2 + bt = -t^2+ 5/2 t\[\]
\[y=c_1+c_2e^{-2x}-t^2+\frac 52 t \]