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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 ?

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# Homework Help: Non-homogenous differential equation

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