Use the power series method to solve the following differential Equation
(2? +1)/" + xy' - y=0)
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Let y=∑aᵣxʳ, then:

y'=∑raᵣxʳ⁻¹ and y''=∑r(r-1)aᵣxʳ⁻², so the DE can be written:


Since we have a 2nd order DE we can prepare two arbitrary constants A and B, which can be allocated as required.



We can collect xʳ terms together:



Then, by replacing r with r+2 in the second summation, we get:





The coefficients of xʳ must all be zero (because this sum must be zero for all x), so we can find what aᵣ must be to satisfy this requirement.

Let r=0 (for the constant term):

-a₀+2a₂=0, a₂=a₀/2. We can use constant A instead of a₀, so a₂=A/2.

When r=1, 6a₃=0, and a₃=0. We can’t determine a₁ so we’ll use constant B in place of a₁.

We have an iterative equation for the a’s:


This can be rewritten: (r-3)aᵣ₋₂+raᵣ=0,


If r=2, a₂=a₀/2=A/2, which we established earlier.

If r=4, a₄=-a₂/4=-A/8.

When r=3, a₃=0, so a₅=a₇=...=0.

When r=6, a₆=-3a₄/6=A/16,


r₁₀=-7a₈/10=7A/256, and so on.

Therefore y=A+Bx+Ax²/2-Ax⁴/8+Ax⁶/16-5Ax⁸/128+7Ax¹⁰/256...,





Now let’s assemble the DE:







You can see that each coefficient is zero.

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