Find two linearly independent power series solutions to the diffrential equation. State the first 3 therms of each series: 2x^2y''+xy'-(1+2x^2)y=0

To find the power series solutions to the given differential equation, we will assume that the solutions can be expressed as power series of the form:

y(x) = ∑(n=0)^(∞) [a_n * x^n]

where a_n are the coefficients to be determined and x is the independent variable.

First, let's find the power series solution in the form of y(x) = ∑(n=0)^(∞) [a_n * x^n]. We substitute this expression into the given differential equation:

2x^2y'' + xy' - (1 + 2x^2)y = 0

Now, let's find the derivatives of y(x) with respect to x:

y' = ∑(n=0)^(∞) [a_n * n * x^(n-1)]
y'' = ∑(n=0)^(∞) [a_n * n * (n-1) * x^(n-2)]

Substituting these derivatives and the power series expression for y(x) back into the differential equation, we have:

2x^2 * (∑(n=0)^(∞) [a_n * n * (n-1) * x^(n-2)]) + x * (∑(n=0)^(∞) [a_n * n * x^(n-1)]) - (1 + 2x^2) * (∑(n=0)^(∞) [a_n * x^n]) = 0

Simplifying this equation, we can group the terms with the same powers of x. Let's consider the terms with x^n:

∑(n=0)^(∞) [(2 * n * (n - 1) * a_n * x^n) + (n * a_n * x^n) + (-a_n * x^n) + (-2 * a_n * x^(n+2))] = 0

To satisfy this equation for every power of x, we set each coefficient of x^n equal to zero. Let's do this for the first few terms:

For n = 0: (2 * 0 * (-1) * a_0) + (0 * a_0) + (-a_0) + (-2 * a_0 * 0) = 0
Simplifying: -a_0 = 0 → a_0 = 0

For n = 1: (2 * 1 * 0 * a_1) + (1 * a_1) + (-a_1) + (-2 * a_1 * 0) = 0
Simplifying: a_1 - a_1 = 0 → a_1 = 0

For n = 2: (2 * 2 * 1 * a_2) + (2 * a_2) + (-a_2) + (-2 * a_2 * 2) = 0
Simplifying: 4a_2 + a_2 - 4a_2 = 0 → a_2 + a_2 = 0 → 2a_2 = 0 → a_2 = 0

For n ≥ 3: (2 * n * (n - 1) * a_n) + (n * a_n) + (-a_n) + (-2 * a_n * (n + 2)) = 0

The coefficients a_0, a_1, and a_2 are all zero. Now we can express a_n in terms of a_(n-3) for n ≥ 3:

2n(n-1)a_n + na_n - a_n - 2(n+2)a_n-2 = 0

Simplifying, we have:

2n(n-1)a_n + na_n = 2(n+2)a_n-2

Dividing by 2 and factoring a_n, we get:

n(n - 1 + 1/2)a_n = (n + 2)a_n-2

Simplifying further, we have:

n^2a_n = a_n-2(n + 2)

Rearranging this equation gives:

a_n = (n + 2)a_n-2 / n^2

Therefore, the coefficients a_n can be determined in terms of a_(n-2). We can start with a_0 = 0, a_1 = 0, and a_2 = 0, and then use the above formula to determine the subsequent coefficients.

Now, let's find the power series solutions by explicitly calculating the first few terms:

For the first solution, let's express y(x) as a power series of the form y(x) = ∑(n=0)^(∞) [a_n * x^n], where a_0 = 0, a_1 = 0, and a_2 = 0.

Therefore, the first three terms of the first power series solution are:

y_1(x) = a_3 * x^3 + a_4 * x^4 + a_5 * x^5 + ...

For the second solution, let's express y(x) as a power series of the form y(x) = ∑(n=0)^(∞) [b_n * x^n], where b_0 = 1, b_1 = 0, and b_2 = 0.

Therefore, the first three terms of the second power series solution are:

y_2(x) = x^2 + b_3 * x^3 + b_4 * x^4 + ...

To find the subsequent coefficients a_n and b_n, we can use the recurrence relation derived earlier:

a_n = (n + 2)a_n-2 / n^2
b_n = (n + 2)b_n-2 / n^2

These equations will allow us to determine the coefficients and thus expand the power series solutions to any desired accuracy.