We are tasked with solving the second-order differential equation using the reduction of order method. The given differential equation is:

$t^2 y'' - t(t + 2)y' + (t + 2)y = 0$ where $y_1 = t$ is already provided as one solution.

Step 1: General form of the reduction of order

The method of reduction of order assumes that the second solution $y_2$ can be written as:

$y_2 = v(t) y_1 = v(t) t$ where $v(t)$ is a function to be determined.

Step 2: Compute the derivatives of $y_2$

We need to find the first and second derivatives of $y_2 = v(t)t$.

1. First derivative: $y_2' = v'(t) t + v(t) \cdot 1 = v'(t) t + v(t)$

2. Second derivative: $y_2'' = v''(t) t + v'(t) \cdot 1 + v'(t) \cdot 1 = v''(t) t + 2v'(t)$

Step 3: Substitute into the original equation

Substituting $y_2 = v(t) t$, $y_2'$, and $y_2''$ into the original differential equation:

$t^2 y_2'' - t(t + 2)y_2' + (t + 2)y_2 = 0$

Substitute the expressions for $y_2$, $y_2'$, and $y_2''$:

$t^2 \left( v''(t) t + 2 v'(t) \right) - t(t + 2)\left( v'(t) t + v(t) \right) + (t + 2) v(t) t = 0$

Step 4: Simplify the equation

Let's expand and simplify the terms step by step.

1. Expanding the first term: $t^2 \left( v''(t) t + 2 v'(t) \right) = t^3 v''(t) + 2 t^2 v'(t)$

2. Expanding the second term: $- t(t + 2)\left( v'(t) t + v(t) \right) = -t(t + 2) v'(t) t - t(t + 2)v(t) = -t^2(t + 2) v'(t) - t(t + 2) v(t)$

Expanding further: $- t^2(t + 2) v'(t) = -t^3 v'(t) - 2 t^2 v'(t)$ $- t(t + 2) v(t) = -t^2 v(t) - 2 t v(t)$

3. Expanding the third term: $(t + 2) v(t) t = t^2 v(t) + 2 t v(t)$

Step 5: Combine like terms

Now, let's combine all the terms:

$t^3 v''(t) + 2 t^2 v'(t) - (t^3 v'(t) + 2 t^2 v'(t)) - (t^2 v(t) + 2 t v(t)) + (t^2 v(t) + 2 t v(t)) = 0$

After simplifying, many terms cancel out: $t^3 v''(t) = 0$

Step 6: Solve for $v(t)$

From the simplified equation $t^3 v''(t) = 0$, we obtain:

$v''(t) = 0$

Integrating twice:

1. First integration gives $v'(t) = C_1$.
2. Second integration gives $v(t) = C_1 t + C_2$.

Step 7: General solution

The second solution $y_2$ is given by:

$y_2 = v(t) t = (C_1 t + C_2) t = C_1 t^2 + C_2 t$

Since $y_1 = t$, the general solution to the differential equation is:

$y(t) = C_1 t^2 + C_2 t$

Final Answer:

The general solution to the differential equation $t^2 y'' - t(t + 2)y' + (t + 2)y = 0$ is:

$y(t) = C_1 t^2 + C_2 t$

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Would you like more details on any part of this solution? Here are some related questions to deepen your understanding:

1. How do we determine the form of $y_2$ in the reduction of order method?
2. Why does $v''(t) = 0$ lead to a linear solution for $v(t)$?
3. Can the reduction of order method be used if $y_1$ is not a simple linear function like $t$?
4. How does the method of variation of parameters compare to reduction of order?
5. How would the solution change if the differential equation had constant coefficients?

Tip: The reduction of order method works well when one solution is already known, and it reduces a second-order differential equation to a first-order problem.