Can you, please, help me to solve this equation:


Here, for me the problem is, I am finding relationship among 3 members: $a_n, a_{n+1}, a_{n+2}$, not between 2 members: $a_n$ and $a_{n+1}$

I would like to solve it using the Frobenius method.



First of all set $(x+1)=t$ to have $t^2y''+ty'-y=0, ~~t\neq -1$ instead. Then, solve the auxiliary equation $$am^2+(b-a)m+c=0$$ wherein

$a=1$ (the coffecient of $t^2$),

$b=+1$ (the cofficient of $t$ above) and $c=-1$

for finding the possible $m$'s.

If $m_1,m_2$ are distinct solutions so the general solution of your Cauchy-Euler ODE will be as $$y_c=C_1t^{m_1}+C_2t^{m_2}$$ If you have $m_1=m_2=m=\frac{a-b}{2a}$ then $y_c=C_1t^m+C_2t^m\ln(t)$ and finally if you have $m=\alpha\pm i\beta$ then $$y_c=t^{\alpha}(C_1\cos(\beta\ln t)+C_2\sin(\beta\ln t))$$ where $C_1,C_2$ are constants. Note that you assume $x+1=t$ before and ofcourse $t\in(0,+\infty)$.

  • $\begingroup$ Thank you. It would be great to see Frobenius method as well. $\endgroup$
    – Bek Abdik
    Apr 19 '13 at 7:32
  • $\begingroup$ + Nice work! How is my dear friend today? $\endgroup$
    – amWhy
    Apr 19 '13 at 18:39
  • $\begingroup$ @amWhy: 50/50. Are you felling better? I have prayed for you up to now. I am working on a messy calculations in a semigroup and I haven't got my desirable results. :-( $\endgroup$
    – Mikasa
    Apr 19 '13 at 18:42
  • $\begingroup$ @BekAbdik: Sorry my friend if I couldn't do what you wanted on time. $\endgroup$
    – Mikasa
    Apr 19 '13 at 18:44
  • $\begingroup$ @amWhy: Thanks Amy. You have done it for me :-) $\endgroup$
    – Mikasa
    Apr 19 '13 at 18:48

Let $y=\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}$ ,

Then $y'=\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}$


$\therefore(x+1)^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r-2}+(x+1)\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r-1}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$

$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_n(x+1)^{n+r}+\sum\limits_{n=0}^\infty(n+r)a_n(x+1)^{n+r}-\sum\limits_{n=0}^\infty a_n(x+1)^{n+r}=0$


Since there are not any indical equations present, that means $r$ can be chosen as any complex number.

However, in fact, take $r=1$ or $r=-1$ will bring the above equation most simplified.

Moreover, in fact, we can find all groups of the linearly independent solutions by just taking $r=-1$ :


$\sum\limits_{n=0}^\infty n(n-2)a_n(x+1)^{n-1}=0$

$\therefore n(n-2)a_n=0$

$\therefore\begin{cases}a_0=a_0\\a_2=a_2\\a_n=0~\forall n\in\mathbb{N}\setminus\{2\}\end{cases}$

Hence $y=\dfrac{C_1}{x+1}+C_2(x+1)$


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