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!