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

To solve the initial value problem:

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

### Step 1: Find the general solution.

We integrate both sides of the equation with respect to \( t \):

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

### Step 2: Use substitution.

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

\[
du = 14t \, dt
\]

This means:

\[
t \, dt = \frac{1}{14} \, du
\]

Substituting into the integral:

\[
s(t) = \int 28 \times \frac{1}{14} \cdot u^3 \, du
\]

Simplifying:

\[
s(t) = 2 \int u^3 \, du
\]

### Step 3: Integrate.

The integral of \( u^3 \) is:

\[
s(t) = 2 \cdot \frac{u^4}{4} + C = \frac{u^4}{2} + C
\]

Substituting back \( u = 7t^2 - 5 \):

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

### Step 4: Apply the initial condition \( s(1) = 1 \).

Substitute \( t = 1 \) into the equation:

\[
1 = \frac{(7(1)^2 - 5)^4}{2} + C
\]

Simplifying:

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

\[
1 = \frac{2^4}{2} + C
\]

\[
1 = \frac{16}{2} + C
\]

\[
1 = 8 + C
\]

So:

\[
C = 1 - 8 = -7
\]

### Final Solution:

The solution to the initial value problem is:

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

---

Would you like any further details on this solution? 

Here are 5 related questions for further exploration:

1. How do you handle a similar initial value problem with different exponents?
2. What other substitution methods could simplify such integrals?
3. How would the solution change with a different initial condition?
4. How can we verify the solution by differentiating \( s(t) \)?
5. How would this approach change if the differential equation were non-linear?

**Tip**: Always recheck your substitutions, especially when using them in integration!

Explore the detailed solution to the initial value problem ds/dt = 28t(7t^2 - 5)^3 with an initial condition s(1) = 1. Understand the steps involving differential equations, integration by substitution, and verification techniques. Suitable for advanced high school and college students.

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