0
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

Hint:

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

$x^2\sum_0^{\infty}(n+\rho)(n+\rho-1)c_nx^{n+\rho-2}-x^2\sum_0^{\infty}(n+\rho)c_nx^{n+\rho-1}+\frac{5}{2}x\sum_0^{\infty}(n+\rho)c_nx^{n+\rho-1}+\frac{1}{2}\sum_0^{\infty}c_nx^{n+\rho}=0$

$\sum_0^{\infty}(n+\rho)(n+\rho-1)c_nx^{n+\rho}-\sum_0^{\infty}(n+\rho)c_nx^{n+\rho+1}+\frac{5}{2}\sum_0^{\infty}(n+\rho)c_nx^{n+\rho}+\frac{1}{2}\sum_0^{\infty}c_nx^{n+\rho}=0$

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 $y= \sum_1^{\infty}c_nx^{n-\frac{1}{2}}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.