Find the second linearly independent solution to the equation

$\displaystyle 2xy''+(5-2x)y'+\frac{y}{x}=0$

using the series method at the point $x=0$.

I try to solve with the following We need to find also y , y’ , y’’.

y=∑_(n=0)^∞▒〖c_n x^n 〗 ,y^'=∑_(n=1)^∞▒〖c_n nx^(n-1) 〗 , y^''=∑_(n=2)^∞▒〖c_n n(n-1)x^(n-2) 〗

We make the usual substitution of the power series. This results in the equation

2∑_(n=2)^∞▒〖c_n n(n-1)x^(n-1) 〗+5∑_(n=1)^∞▒〖c_n nx^(n-1) 〗-2∑_(n=1)^∞▒〖c_n nx^n 〗+∑_(n=0)^∞▒〖c_n x^(n-1) 〗=0 We can start the second sum at n = 0 without changing anything else. To make each term include x^n in its general term, we shift the index of summation in the first sum by +1 (replace n with n + 1)

2∑(n=1)^∞▒〖c(n+1) n(n+1) x^n 〗+5∑(n=0)^∞▒〖c(n+1) (n+1) x^n 〗-2∑_(n=1)^∞▒〖c_n nx^n 〗+∑(n=1)^∞▒〖c(n+1) x^n 〗=0

When n =1, 4c_2+10c_2-2c_1+c_2=0 , 15c_2-2c_1=0 , c_2=(2c_1)/15 〖2c〗(n+1) n(n+1)+5c(n+1) (n+1)-2c_n n+c_(n+1)=0

c_(n+1) (2n(n+1)+5(n+1)+1)-2c_n n=0 , c_(n+1)=(2c_n n)/(2n(n+1)+5(n+1)+1)

I'm stuck here what next step .

  • $\begingroup$ You now have something of the form $\sum_{n=-1}^{\infty} b_nx^n = 0$. So you should now make sure each of the $b_i$ is 0. $\endgroup$ – Aahz Jun 27 '14 at 15:13
  • $\begingroup$ First off you need to shift the index so that you have terms of the form $$\sum_{n=0}^{\infty}b_nx^n$$ $\endgroup$ – Chinny84 Jun 27 '14 at 15:46
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    $\begingroup$ Your question was edited twice and now you undid all the changes (why?). Please edit the question again to make it more readable. $\endgroup$ – Ludolila Jun 27 '14 at 18:56

I would say that the series of $\sum_0^\infty c_n x^n \,$ not solve the equation.


$x^2y''-(x^2-\frac{5}{2}x)y'+\frac{y}{2}=0,\,\Rightarrow \color{red}{y= \sum_0^{\infty}c_nx^{n+\rho}},\,y' = \sum_0^{\infty}(n+\rho)c_nx^{n+\rho-1},\,y'' = \sum_0^{\infty}(n+\rho)(n+\rho-1)c_nx^{n+\rho-2}$



Determine $ \rho:\,\, n=0\,\,\Rightarrow \rho(\rho-1)c_0x^{\rho}+\frac{5}{2}\rho c_0 x^{\rho}+\frac{1}{2}c_0x^{\rho}=\frac{1}{2}c_0x^{\rho}(2\rho^2+3\rho+1)=0$

$\Rightarrow 2\rho^2+3\rho+1=0 \Rightarrow \rho_{1}=-1,\rho_{2}=-\frac{1}{2};\,\,c_0 = $any constant

First solving equations by substitution series $y= \sum_1^{\infty}c_nx^{n-1},$ second solving by substitution $y= \sum_1^{\infty}c_nx^{n-\frac{1}{2}}.$


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