Solve the initial value problem ds/dt=28t(7t^2-5)^3, s(1)=1

To solve the initial value problem:

\[
\frac{ds}{dt} = 28t(7t^2 - 5)^3, \quad s(1) = 1
\]

### Step 1: Integrate the differential equation.

We begin by finding the general solution to the differential equation by integrating both sides with respect to \( t \):

\[
s(t) = \int 28t(7t^2 - 5)^3 \, dt
\]

### Step 2: Apply substitution.

Let \( u = 7t^2 - 5 \). Then, the derivative \( du \) with respect to \( t \) is:

\[
du = 14t \, dt
\]

Thus, the integral becomes:

\[
s(t) = \int 2(7t^2 - 5)^3 \, du
\]

Now, integrate:

\[
s(t) = \frac{(7t^2-5)^4}{2} + C
\]

Finally, apply initial condition \(s(1) = 1\):

\[
1 = s(1) 
\]


### Final Answer:
The final answer is . 


Learn how to solve the initial value problem ds/dt = 28t(7t^2 - 5)^3 with a step-by-step solution. This problem involves differential equations, integration, and substitution techniques. Suitable for advanced high school or college students studying calculus.

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