Math: Differential Equations

I'm given y''= sin(x) with the initial conditions of y(0)=0 and y'(0)=2. I've already taken the integral two times to get an answer of y= -sin(x)+3x, however, I'm stuck as to how to make this equation in terms of x=...

Please explain, I'd really appreciate it!

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  1. if y'' = sinx
    then y' = -cosx + c
    given : 2= -cos0 + c
    2 = -1 + c
    c = 3

    y' = -cosx + 3
    y = -sinx + 3x + k
    given: 0 = =sin0 + 0 + k
    0 = 0 + 0 + k --> k = 0

    y = -sinx + 3x
    you had that, you are done!
    you have it expressed in terms of x

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    posted by Reiny

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