Question

Solve the initial value problem.
ds 36t(91²-7)³, s(1)=1
dt
The solution is s =

Answer 1

Integrate both sides.

`\displaystyle \int (ds)/(dt) \, dt = \int 36t (9t^2 - 7)^3 \, dt`

On the right side, substitute
`u=9t^2-7` and
`du=18t\,dt`, so that

`\displaystyle \int 36t (9t^2 - 7)^3 \, dt = 2 \int u^3 \, du = \frac12 u^4 + C = \frac12 (9t^2 - 7)^4 + C`

Use the initial value to solve for
`C`.

`s(1) = 1 \implies 1 = \frac12 (9 - 7)^4 + C \implies 1 = 8 + C \implies C=-7`

Then the particular solution is

`s(t) = \boxed{\frac12 (9t^2 - 7)^4 - 7}`