Use power series to solve the initial value problem.

y double prime plus xy prime minus 2 y equals 0y′′+xy′−2y=0​; 

y left parenthesis 0 right parenthesis equals 3y(0)=3​, 

y prime left parenthesis 0 right parenthesis equals 0y′(0)=0

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Part 1

y left parenthesis x right parenthesis equalsy(x)=enter your response here

To solve the initial value problem \( y'' + xy' - 2y = 0 \) with the initial conditions \( y(0) = 3 \) and \( y'(0) = 0 \) using power series, we assume a solution of the form:

\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]

First, we differentiate this series to find \( y'(x) \) and \( y''(x) \):

\[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \]
\[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]

Substituting \( y \), \( y' \), and \( y'' \) into the differential equation:

\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + x \sum_{n=1}^{\infty} n a_n x^{n-1} - 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]

Simplifying the second term:

\[ x \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=1}^{\infty} n a_n x^n \]

Combining all the sums:

\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + \sum_{n=1}^{\infty} n a_n x^n - 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]

We shift the indices to match the powers of \( x \):

\[ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + \sum_{n=1}^{\infty} n a_n x^n - 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]

Now, separating the first term, we get:

\[ (a_2 \cdot 2 \cdot 1) + \sum_{n=1}^{\infty} ((n+2)(n+1) a_{n+2} + n a_n - 2 a_n) x^n = 0 \]

Thus, the coefficient of each \( x^n \) must be zero, giving the recurrence relation:

\[ a_2 \cdot 2 \cdot 1 = 0 \implies a_2 = 0 \]
\[ (n+2)(n+1) a_{n+2} + (n - 2) a_n = 0 \]

Simplifying the recurrence relation:

\[ a_{n+2} = \frac{2 - n}{(n+2)(n+1)} a_n \]

Using the initial conditions \( y(0) = 3 \) and \( y'(0) = 0 \):

\[ a_0 = 3 \]
\[ a_1 = 0 \]

Now we compute the coefficients \( a_n \):

For \( n = 0 \):

\[ a_2 = \frac{2 - 0}{2 \cdot 1} a_0 = 0 \]

For \( n = 1 \):

\[ a_3 = \frac{2 - 1}{3 \cdot 2} a_1 = 0 \]

For \( n = 2 \):

\[ a_4 = \frac{2 - 2}{4 \cdot 3} a_2 = 0 \]

For \( n = 3 \):

\[ a_5 = \frac{2 - 3}{5 \cdot 4} a_3 = 0 \]

For \( n = 4 \):

\[ a_6 = \frac{2 - 4}{6 \cdot 5} a_4 = 0 \]

Continuing in this manner, we see that all coefficients \( a_n \) for \( n \geq 2 \) are zero. Hence, the power series solution is:

\[ y(x) = a_0 + a_1 x = 3 \]

So, the solution to the initial value problem is:

\[ y(x) = 3 \]

Would you like more details or have any questions? Here are some related questions:

1. How do you determine the general term for the power series solution?
2. Can you provide another example of solving an initial value problem using power series?
3. How do the initial conditions affect the coefficients in the power series?
4. What are the steps to simplify the recurrence relation?
5. How can you verify the solution of the differential equation using power series?
6. What if the initial conditions were different, how would that change the solution?
7. Can you explain the process of differentiating the power series in more detail?
8. What is the significance of the power series representation in solving differential equations?

**Tip:** Always check if the initial conditions lead to simplifications in the power series coefficients, which can make solving the problem easier.

Explore how power series are used to solve differential equations, specifically the initial value problem y'' + xy' - 2y = 0 with initial conditions y(0) = 3 and y'(0) = 0. Learn step-by-step how to derive the power series solution and its coefficients.

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