Find two linearly independent Frobenius series solutions of : x 2 y\'\' - xy\' +

Question

Find two linearly independent Frobenius series solutions of :

x2y'' - xy' + (x2 + 1)y = 0

For each Frobenius solution, find the first four non-zero terms.

[please try to present solution like what is shown here]

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Explanation / Answer

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

2xy + x^2(dy/dx) - [(1)y + x(dy/dx)] + 2xy + x^2(dy/dx) + (dy/dx) = 0

4xy + 2x^2(dy/dx) - y - x(dy/dx) + (dy/dx) = 0

2x^2(dy/dx) - x(dy/dx) + (dy/dx) = y - 4xy

(dy/dx)[2x^2 - x + 1] = y(1-4x)

dy/y = [(1 - 4x)/(2x^2 - x + 1)](dx) let (2x^2 - x + 1) = u

Integrate the both sides of equation (4x - 1)dx = du

lny = [-1(4x - 1)/u](du/(4x - 1)) dx = du/(4x - 1)

After simplification

lny = -du/u

lny = -lnu

therefore

lny = -ln(2x^2 - x + 1)

lny = ln[1/(2x^2 - x + 1)]

Taking anti-log on both sides

y = [1/(2x^2 - x + 1)]

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