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 power series solutions to the given differential equation, we can assume that the solution can be expressed as a power series of the form:
y(x) = Σₖ aₖxₖ,
where aₖ represents the coefficients of the power series and k is a non-negative integer.
To begin, let's find the first few terms of the power series solutions.
Firstly, let's find the power series solution around the point x = 0.
Differentiating the power series with respect to x, we have:
y'(x) = Σₖ aₖkx^(k-1).
Differentiating again, we get:
y''(x) = Σₖ aₖ(k)(k-1)x^(k-2).
Now, plug these expressions for y, y', and y'' back into the differential equation:
2x²(Σₖ aₖ(k)(k-1)x^(k-2)) + x(Σₖ aₖkx^(k-1)) - (1 + 2x²)(Σₖ aₖxₖ) = 0.
Let's re-arrange the terms:
Σₖ 2aₖ(k)(k-1)x^k + Σₖ aₖkx^k - Σₖ aₖx^k - Σₖ 2aₖx^(k+2) - Σₖ aₖx^k = 0.
Now, group the terms with the same powers of x:
Σₖ (2aₖ(k)(k-1) + aₖk - aₖ)x^k - Σₖ 2aₖx^(k+2) = 0.
For this equation to hold for all values of x, each coefficient of the corresponding powers of x must be zero.
Considering the coefficient of x^0:
(2a₀(0)(0-1) + a₀(0) - a₀) = 0,
-a₀ = 0,
a₀ = 0.
Now, let's find the coefficients for x^1:
(2a₁(1)(1-1) + a₁(1) - a₁) = 0,
a₁ = a₁.
Since a₁ can be any value, let's choose a₁ = 1 for simplicity.
Moving on to the coefficients for x^2:
(2a₂(2)(2-1) + a₂(2) - a₂) = 0,
2a₂ + 2a₂ = 0,
4a₂ = 0,
a₂ = 0.
Now that we have determined a₀ = 0, a₁ = 1, and a₂ = 0, we can substitute these values back into the power series expression for y(x):
y(x) = a₀x⁰ + a₁x¹ + a₂x² + Σₖ aₖxₖ
= 0 + 1x + 0x² + Σₖ aₖxₖ
= x + Σₖ aₖxₖ.
Therefore, one power series solution is y₁(x) = x.
For the second linearly independent solution, we can use the technique of Frobenius series method. Let's assume the second power series solution is:
y₂(x) = cxz(x),
where z(x) = ∞∑ₖ₌₀ bₖxⁿ and c is a constant to be determined.
Differentiating y₂(x) once and twice, we get:
y₂'(x) = cxz'(x) + cx'z(x),
y₂''(x) = cxz''(x) + 2cx'z'(x) + cx''z(x).
Now, substitute these expressions back into the differential equation:
2x²(cxz''(x) + 2cx'z'(x) + cx''z(x)) + x(cxz'(x) + cx'z(x)) - (1 + 2x²)(cxz(x)) = 0.
Expand and simplify:
2cx²z''(x) + 4cxz'(x) + 2cx²z'(x) + cx''z(x) + cxz(x) - cx²z(x) - 2x²cxz(x) = 0.
Factor out the common terms:
cxz(x)(2x²z''(x) + 2xz'(x) + cx''(x) + 2x² - c - 1) = 0.
For this equation to hold, either cxz(x) = 0 (which we will disregard) or the expression in parentheses must equal zero:
2x²z''(x) + 2xz'(x) + cx''(x) + 2x² - c - 1 = 0.
We need to find z(x) and c that satisfy this equation. One possible choice for z(x) is the geometric series:
z(x) = 1 + Σₖ₌₁∞ bₖxⁿ,
where bₖ = (k-1) / (2k(k+1)).
Since z(x) is a power series with a non-zero constant term, we can let c = 1. Thus, our second power series solution becomes:
y₂(x) = xz(x)
= x(1 + Σₖ₌₁∞ bₖxⁿ).
Let's find the first three terms of y₂(x):
y₂(x) = x(1 + Σₖ₌₁∞ bₖxⁿ)
= x(1 + ∑ₖ₌₁∞ (k-1) / (2k(k+1))xⁿ)
= x + Σₖ₌₂∞ (k-1) / (2k(k+1))xⁿ.
Therefore, the second power series solution is y₂(x) = x + Σₖ₌₂∞ (k-1) / (2k(k+1))xⁿ.