Solve the initial value problem.

= 36(9t^{2}-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 = 9t^{2} -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(9t^{2} – 7)^{3}dt = ∫2u^{3} du = u^{4}/2 +C

Replacing u by 9t^{2}-7, s = ∫36(9t^{2}-7)^{3}dt = (9t^{2}-7)^{4}+C

Using s(1) = 9, solve for C.

s = (9t^{2}-7)^{4} +C

9 = (9(1)^{2} – 7)^{4}+C

C = 1

Thus , s = (9t^{2} – 7 )^{4} + 1.