Math: Differential Equations
posted by Student .
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!

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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