Solve the following initial value problems! Help Me !?

Question

hey guys would really appreciate it if can show me the working in details for this questions.. thanks =)

Solve the following initial value problems

ds/dt = 12t(3t^2-1)^3 , s(1) = 3

and the answer is : s = 1/2 (3t^2 - 1)^4 - 5

Answer 1

Given ds/dt = 12t(3t^2-1)^3

  s = ∫12t(3t^2-1)^3 dt

  

Let u = 3t^2-1

   du/dt = 6t

s = ∫12t(3t^2-1)^3 dt

s = 2∫6t(3t^2-1)^3 dt

s = 2∫(du/dt)*(u)^3 dt

s = 2∫(u)^3 du

s = 2[(u)^4/4]+C

s = (u)^4/2 + C

s = [(3t^2-1)^4]/2 + C

When t=1, s=3,

3 = [(3-1)^4]/2 + C

3 = 16/2+c

3 = 8 + C

C = 8-3

C = 5

So s = [(3t^2-1)^4]/2 + 5

or

s = 1/2 (3t^2 - 1)^4 - 5.

Hope this helps

Answer 2

ds/dt   = 12t(3t^2 -1)^3

ds =  12t(3t^2 -1)^3 dt

integrating

s  = integral 12t(3t^2 - 1)^3

put x = 3t^2 - 1

    dx =  6t dt

       = 2x^3

integrating  =   2x^4/4 = 1/2 x^4+c

s = 1/2 (3t^2 -1 )^4+c

since s(1) is = 3  we  get

s(1) = 3 = 1/2(3*1-1)^4 +c

           3 =  8+c

           c = -5

so

  s = 1/2 (3t^2 - 1)^4 - 5