We are given the following differential equation,
$$x \cdot \frac{d^2y}{dx^2} + (2+x)\frac{dy}{dx}+y=0$$
i) Show the following,
$$x \cdot \frac{d^2y}{dx^2} + (2+x)\frac{dy}{dx}+y=\sum{[(n+1)c_n+(n+1)(n+2)c_{n+1}]x^n}$$
ii) Explain that if the power series $\sum{c_nx^n}$ satisfies the differential equation, then
$$c_{n+1}=-\frac{1}{n+2}c_n$$
iii) Determine using the solution from (ii), the radius of convergence for an arbitrary power series solution to the differential equation. [Hint: Hint: use ratio test to examine cn + 1 / cn for n → ∞. It is not necessary to calculate the coefficients $c_n$.]
My Working
P.s the riemann sum is from n=0 to infinity in part (i) after the equalsign.
For i)
I know that I have to use the power series, $$y=\sum _{ n=0 }^{ \infty }{c_n·x^n } $$
Therefore, $$\frac{dy}{dx}=\sum _{ n=1 }^{ \infty }{nc_n·x^{n-1} }$$ $$\frac{d^2y}{dx^2}=\sum _{ n=2 }^{ \infty }{n(n-1)c_nx^{n-2} }$$
Now it is only the solution, if $$x \cdot \frac{d^2y}{dx^2} + (2+x)\frac{dy}{dx}+y=0$$ $$x \sum _{ n=2 }^{ \infty }{n(n-1)c_nx^{n-2} } + (2+x)\sum _{ n=1 }^{ \infty }{nc_n·x^{n-1} }+\sum _{ n=0 }^{ \infty }{c_n·x^n }=0$$
Which becomes, $$\sum _{ n=2 }^{ \infty }{n(n-1)c_nx^{n-1} } + \sum _{ n=1 }^{ \infty }{2nc_n·x^{n-1} }+\sum _{ n=1 }^{ \infty }{nc_n·x^{n} }+\sum _{ n=0 }^{ \infty }{c_n·x^n }=0$$
Then I changed the limits on my sums so that everything is starting from n=0, and I ended up with the following expression,
$$\sum _{ n=2 }^{ \infty }{n(n-1)c_nx^{n-1} } + \sum _{ n=1 }^{ \infty }{2nc_n·x^{n-1} }+\sum _{ n=1 }^{ \infty }{nc_n·x^{n} }+\sum _{ n=0 }^{ \infty }{c_n·x^n }=0$$
$$3c_1 +\sum _{ n=0 }^{ \infty }{ [(n+2)(n+1) c_{n+2} x^{n+1}+2c_{n+2} (n+2) x^{n+1}+(n+2) c_{n+2} x^{n+2}+c_n x^n } $$
Where ofcourse $c_1=0$ and now I can not factorise it either, because the power on x is not the same and I can not understand how to get to the answer.
Part ii)
I have solved part (ii) by saying the following, $$(n+1)c_n+(n+1)(n+2)c_{n+1}=0$$
Hence solving we come cross to the following,
$$c_{n+1}=-\frac{1}{n+2}c_n$$
Part iii)
This part I do not understand. Isnt convergence the limit value for $x$? Meaning that the convergence radius is just the value such that, $$ x∈[-\rho, \rho]$$
The equation that you have derived while performing the series solution, as you have stated is:
$$\sum_{n=2}^{\infty} n (n-1)c_n x^{n-1} + \sum_{n=1}^{\infty} 2nc_n x^{n-1} + \sum_{n=1}^{\infty} nc_n x^{n} + \sum_{n=0}^{\infty} c_n x^{n}=0$$
Which is exactly the same as the one below:
$$\sum_{n=0}^{\infty} n (n-1)c_n x^{n-1} + \sum_{n=0}^{\infty} 2nc_n x^{n-1} + \sum_{n=0}^{\infty} nc_n x^{n} + \sum_{n=0}^{\infty} c_n x^{n}=0$$
Since the terms you obtain by putting $n=0,1$ in the above series expansions are all zero.
And finally the above set of equation can be re-stated as follows:
$$\sum_{n=0}^{\infty} n (n+1)c_{n+1} x^{n} + \sum_{n=0}^{\infty} 2(n+1)c_{n+1} x^{n} + \sum_{n=0}^{\infty} nc_n x^{n} + \sum_{n=0}^{\infty} c_n x^{n}=0$$
And now you can take $x^n $ common from all the series and find the relation stated in (i).
Again as for the radius of convergence, $$R=\lim_\limits {n\to\infty} |\frac {c_n}{c_{n+1}}|=\lim_\limits {n\to\infty} |(n+2)|= \infty$$