Question

please I need a slution for this : y"-4y'+3y=exp(x)+exp(-x)

Answer 1

what do you get for the homogenous part?

Answer 2

A*exp(4x) + B*exp(x) what I need is the particular solution

Answer 3

your roots are 3 and 1, but ill assume thats a typo

there is a trial and error method; assume a

y = exp(x) + b exp(-x) is a solution

find y' and y'' of those

Answer 4

shoulda been an a in front of the first one

Answer 5

yes yes excuse me it's 3 and 1

Answer 6

y = a exp(x) + b exp(-x)
y' = a exp(x) - b exp(-x)
y'' = a exp(x) + b exp(-x)

1y = a exp(x) + b exp(-x)
-4y' = -4a exp(x) +4b exp(-x)
3y'' = 3a exp(x) + 3b exp(-x)
---------------------------
e(-x)+e(x) = 0a e(x) + 8b e(-x)

b = 1/8 but that 0 has me a little iffied

Answer 7

ok it seems correct

Answer 8

thanks for your help

Answer 9

might have to consider an "ax" instead of an a

y = ax e(x) + b e(-x)
y' = a e(x) + ax e(x) - b e(-x)
y'' = 2a e(x) + ax e(x) + b e(-x)

y = ax e(x) + b e(-x)
-4y' = -4a e(x) -4ax e(x) +4b e(-x)
+3y'' = 6a e(x) + 3x e(x) + 3b e(-x)
---------------------------
= 2a e(x) + 0 +8b e(-x)

b = 1/8\
a = -1/2

Answer 10

yes this is what i found too. Thanks

Answer 11

youre welcome

Answer 12

but I remain unconvinced of the form of y

Answer 13

y = yh + yp

Answer 14

if you know how to play with a wronskian setup


Wx W Wy
e(3x) 0 e(x)
3e(3x) (e(x)+e(-x)) e(x)

Wx = -(e(2x) + 1)
Wy = (e(4x)+e(2x))
W = -2e(4x)