- #1
James889
- 192
- 1
Hi,
My memory ran away, i completely forgot how to do this =/
I need to solve
[tex]y'' + 4y = tsint,~~y(0) = 0,~y'(0) = 1[/tex]
with the general solution [tex]C_1sin(2t) + C_2cos(2t)[/tex]
[tex]y = Atsin(t) +Btcos(t)[/tex]
[tex]y' = Asin(t) + Atcos(t) +Bcos(t) -Btcos(t)[/tex]
[tex]y'' = Acos(t) + Acos(t) -Atsin(t) - Bsin(t) -Bsin(t) -Btsin(t)[/tex]
substitute back into the first equation:
[tex]2Acos(t) - 2Bsin(t) + 3Atsin(t) + 3Btcos(t)[/tex]
So if i let A be [tex]1/3[/tex] i get [tex]2/3cos(t) + tsin(t)[/tex]
How did you determine the [tex]C_1[/tex] and [tex]C_2[/tex] constants ?
My memory ran away, i completely forgot how to do this =/
I need to solve
[tex]y'' + 4y = tsint,~~y(0) = 0,~y'(0) = 1[/tex]
with the general solution [tex]C_1sin(2t) + C_2cos(2t)[/tex]
[tex]y = Atsin(t) +Btcos(t)[/tex]
[tex]y' = Asin(t) + Atcos(t) +Bcos(t) -Btcos(t)[/tex]
[tex]y'' = Acos(t) + Acos(t) -Atsin(t) - Bsin(t) -Bsin(t) -Btsin(t)[/tex]
substitute back into the first equation:
[tex]2Acos(t) - 2Bsin(t) + 3Atsin(t) + 3Btcos(t)[/tex]
So if i let A be [tex]1/3[/tex] i get [tex]2/3cos(t) + tsin(t)[/tex]
How did you determine the [tex]C_1[/tex] and [tex]C_2[/tex] constants ?