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.