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Solve the differential equation $(4-x^2) y'' + 2y = 0$ by means of a power series about the point $x=0$.

a. Find the recurrence solution

b. Find the first four terms in each of two linearly independent solutions

c. Find the general term in each solution

d. Solve the initial value problem: $y(0)=0, y'(0)=1$

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  • $\begingroup$ This ODE is easy to solve. $y(x)=(x^2-4)f(x)$ leads to a separable ODE. $\endgroup$ – JJacquelin Nov 8 '14 at 21:27
  • $\begingroup$ @JJacquelin, does it? $\endgroup$ – Simon S Nov 9 '14 at 0:02
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This is not the response to the question initially raised. This is the response to Simon S , showing how to find the general solution of the ODE, thanks to a simple change of function. $$(4-x^2) y'' + 2y = 0$$ $$y=x^2f(x)$$ $$y'=2xf+x^2f'$$ $$y''=2f+4xf'+x^2f''$$ $$(4-x^2)\big(2f+4xf'+x^2f''\big) + 2x^2f = 0$$ After simplification : $$2f+(4x-x^3)f' = 0$$ $$\frac{f'}{f}=\frac{2}{x^3-4x}$$ $$\ln\mid f\mid =\frac{1}{4}\ln\mid 4-x^2\mid -\frac{1}{2}\ln\mid x\mid +constant$$ $$f=C\frac{\mid 4-x^2\mid ^{1/4}}{\mid x\mid ^{1/2}}$$ $$y=x^2f=C x^{3/2}\mid 4-x^2\mid ^{1/4}$$

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