F''(x)=x^-3/2
f'(4)=1
f(0)=0
how do I solve this differential equation?
f" = x^-(3/2)
f' = -2x^-1/2 + c
f = -2(2)x^1/2 + c x + d
when x = 4
-2(1/2) + c = 1
-1 + c = 1
c = 2
so
f' = -2x^-(1/2) + 2
when x = 0
0 = d
so
f = -4 sqrt(x) + 2 x
check my arithmetic !