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We are given the second-order differential equation:

\[
24t^2 y'' + 8t(t - 6)y' - 8(t - 6)y = 0, \quad t > 0.
\]

Also, it is provided that \( y_1(t) = 2t \) is a solution, and we are tasked to find a second solution \( y_2 \) using the **reduction of order** method. Here are the steps involved:

### Part i) Solve for \( w(t) \)

We will find \( y_2(t) \) in the form:
\[
y_2(t) = v(t)y_1(t),
\]
where \( y_1(t) = 2t \), and \( v(t) \) is a function to be determined. This leads to:

\[
y_2'(t) = v'(t)y_1(t) + v(t)y_1'(t) = v'(t) \cdot 2t + v(t) \cdot 2,
\]
and
\[
y_2''(t) = v''(t) \cdot 2t + 2v'(t) + v'(t) \cdot 2.
\]

Substitute these into the original equation:
\[
24t^2 \left( v''(t) \cdot 2t + 2v'(t) \right) + 8t(t - 6) \left( v'(t) \cdot 2t + v(t) \cdot 2 \right) - 8(t - 6)v(t) \cdot 2t = 0.
\]

Simplify the equation:
- Collect terms for \( v'(t) \) and \( v(t) \).
- Solve the resulting equation to find the form of \( v(t) \).

Now, using **reduction of order** involves assuming:
\[
w = v',
\]
and solving the first-order equation for \( w(t) \).

Let me compute this step for you now.The equation obtained after substitution of \( y_2 = v(t)y_1(t) \) into the original differential equation and simplifying is:

\[
24t^2 \left( 2t v''(t) + 2 v'(t) \right) + 8t(t - 6)\left( 2t v'(t) + 2v(t) \right) - 2t(8t - 48)v(t) = 0.
\]

This can be rewritten to simplify further before solving for \( w(t) \) (where \( w = v' \)).

To proceed with part **i** and solve for \( w(t) \), I'll solve this equation under the assumption \( w'(t) + P(t)w = 0 \). Let me now simplify this equation to solve for \( w(t) \).The simplified form of the equation is:

\[
16t^2 \left( t v'(t) + 3t v''(t) - 3 v'(t) \right) = 0.
\]

Now, we solve this differential equation for \( v(t) \), where \( w(t) = v'(t) \). 

Let me proceed by solving it for \( w(t) \).The solution to the equation is:

\[
v(t) = C_1 + t^{2 - \frac{t}{3}} \left( C_2 \sin\left( \frac{\log(t)}{3} \cdot \left|\text{Im}(t)\right| \right) + C_3 \cos\left( \frac{\log(t)}{3} \cdot \text{Im}(t) \right) \right).
\]

Now that we have \( v(t) \), we can proceed with finding \( w(t) = v'(t) \) and use the given initial condition \( w(1) = -1 \).

Let me compute that now.Given the solution for \( v(t) \), we can compute \( w(t) = v'(t) \) and apply the condition \( w(1) = -1 \).

Let me differentiate \( v(t) \) to find \( w(t) \) and then solve for the constants using the initial condition \( w(1) = -1 \).It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to proceed manually with guidance on how to differentiate \( v(t) \) to find \( w(t) \), or any other part of the solution? Let me know how you'd like to proceed!

Here are some relative questions to deepen your understanding:
1. What is the purpose of the reduction of order method?
2. Why is it important to have a second independent solution for differential equations?
3. How can initial conditions affect the form of the final solution in such problems?
4. Could you explain the relationship between \( w(t) \) and \( v'(t) \)?
5. What methods are commonly used to verify the correctness of the solution for differential equations?

**Tip:** Always substitute the final solutions back into the original differential equation to check their validity!

Consider the differential equation: 24t²y'' + 8t(t - 6)y' - 8(t - 6)y = 0, t > 0. Given that y₁(t) = 2t is a solution, apply the reduction of order method to find another solution y₂ for which y₁ and y₂ form a fundamental solution set.

Learn how to solve the second-order differential equation 24t²y'' + 8t(t - 6)y' - 8(t - 6)y = 0 using the reduction of order method. This example demonstrates how to find a second solution, y₂, when y₁(t) = 2t is given. Ideal for students studying differential equations at the college level.

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