t2y′′ − t(t + 2)y′ + (t + 2)y =0 . Solve using reduction of order method where y1 = t

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:

\[
y(t) = C_1 t^2 + C_2 t
\]

---

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.

Solve using reduction of order method where y1 = t for the differential equation t^2 y'' − t(t + 2)y' + (t + 2)y = 0.

Learn how to solve the second-order differential equation t^2 y'' − t(t + 2)y' + (t + 2)y = 0 using the reduction of order method. This step-by-step guide explains the process, key formulas, and mathematical concepts like linear independence and second-order differential equations. Suitable for undergraduate students or advanced high school calculus learners.

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