# Problem 854

Solve the initial value problem.

= 36(9t2-7)3, s(1) = 9

Solution:-

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

find du.

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.