James889
- 190
- 1
Hi,
My memory ran away, i completely forgot how to do this =/
I need to solve
y'' + 4y = tsint,~~y(0) = 0,~y'(0) = 1
with the general solution C_1sin(2t) + C_2cos(2t)
y = Atsin(t) +Btcos(t)
y' = Asin(t) + Atcos(t) +Bcos(t) -Btcos(t)
y'' = Acos(t) + Acos(t) -Atsin(t) - Bsin(t) -Bsin(t) -Btsin(t)
substitute back into the first equation:
2Acos(t) - 2Bsin(t) + 3Atsin(t) + 3Btcos(t)
So if i let A be 1/3 i get 2/3cos(t) + tsin(t)
How did you determine the C_1 and C_2 constants ?
My memory ran away, i completely forgot how to do this =/
I need to solve
y'' + 4y = tsint,~~y(0) = 0,~y'(0) = 1
with the general solution C_1sin(2t) + C_2cos(2t)
y = Atsin(t) +Btcos(t)
y' = Asin(t) + Atcos(t) +Bcos(t) -Btcos(t)
y'' = Acos(t) + Acos(t) -Atsin(t) - Bsin(t) -Bsin(t) -Btsin(t)
substitute back into the first equation:
2Acos(t) - 2Bsin(t) + 3Atsin(t) + 3Btcos(t)
So if i let A be 1/3 i get 2/3cos(t) + tsin(t)
How did you determine the C_1 and C_2 constants ?