Solve the initial value problem.
= 36(9t2-7)3, s(1) = 9
To solve this initial value problem for s as a function of t, note that
=f(t) → ds = f(t)dt →s = ∫ f(t)dt.
To integrate ∫f(t)dt, use the Substitution Rule.
The Substitution Rule state if u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫f(g(x))g’(x)dx=∫f(u)du.
Substitute u = g(t) and du = g’(t)dt to obtain the integral ∫f(u)du.
Find the function u.
u = 9t2 -7
du = 18t dt
Since du = 18 dt , 2 du = 36t dt.
Rewrite the integral in terms of u and integrate with respect to u.
∫36(9t2 – 7)3dt = ∫2u3 du = u4/2 +C
Replacing u by 9t2-7, s = ∫36(9t2-7)3dt = (9t2-7)4+C
Using s(1) = 9, solve for C.
s = (9t2-7)4 +C
9 = (9(1)2 – 7)4+C
C = 1
Thus , s = (9t2 – 7 )4 + 1.