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