So, this eq. has irregular points at $x=1$ and $x=0$; Using Frobenius method I can expand this thing arround $x=0$ anyway. As long as

$$y(x)=x^\mu\sum_{n=0}^\infty a_nx^n$$

Ok, cool. I won't write down all the steps I did, but I get that $\mu_{1,2}=0,1$. Is that correct? I think so. And also

$$x^\mu\sum_{n=0}^\infty [((n+\mu+2)(n+\mu) + 1)a_n - (n+\mu)(n+\mu+1)a_{n+1}]x^n=0$$


$$((n+\mu+2)(n+\mu) + 1)a_n=(n+\mu)(n+\mu+1)a_{n+1}$$

Alright, when $\mu=0$ I get

$$((n+2)n+1)a_n=n(n+1)a_{n+1}$$ $$(n+1)^2a_n=n(n+1)a_{n+1}$$

And finally


Let $a_0=1$. Then $a_1$ and other coefficients make no sense. What do I do now? Need to solve this.


One have to take care of the first terms of the sum.

In case of $\mu=0$ we have $y=\sum_{n=0}^\infty a_nx^n$

$y'=\sum_{n=1}^\infty n a_nx^{n-1}$

$y''=\sum_{n=2}^\infty n(n-1) a_nx^{n-2}$ $$x(x-1)\sum_{n=2}^\infty n(n-1) a_nx^{n-2}+3x\sum_{n=1}^\infty n a_nx^{n-1}+\sum_{n=0}^\infty a_nx^n=0$$ $$\sum_{n=2}^\infty n(n-1) a_nx^n -\sum_{n=2}^\infty n(n-1) a_nx^{n-1} +3\sum_{n=1}^\infty n a_nx^n+\sum_{n=0}^\infty a_nx^n=0$$ $$\sum_{n=2}^\infty n(n-1) a_nx^n -\sum_{n=1}^\infty (n+1)n a_{n+1}x^n +3\sum_{n=1}^\infty n a_nx^n+\sum_{n=0}^\infty a_nx^n=0$$ $$a_0x^0 -2a_2x+3a_1x+a_1x+ \sum_{n=2}^\infty (n(n-1)a_n-(n+1)na_{n+1}+3na_n+a_n)x^n=0$$ $$a_0x^0 -2a_2x+4a_1x+ \sum_{n=2}^\infty ((n^2+2n+1)a_n-(n+1)na_{n+1})x^n=0$$ This implies : $$a_0=0$$ $$a_2=2 a_1$$ $$a_{n+1}=\frac{n+1}{n}a_n \qquad n\geq 2$$ From this : $$a_n=na_1 \qquad n\geq 1$$ $$y(x)=a_1\sum_{n=1}^\infty nx^n = a_1 \frac{x}{(x-1)^2}$$ This is a first family of solution of the ODE, any coefficient $a_1$.

$$ $$

From the first solution $y=\frac{x}{(x-1)^2}$, a second family of solution can easily be obtained thanks to the reduction of order method with the change of function $y(x)=\frac{x}{(x-1)^2}u(x)$. But this is not the Frobenius method.

If you definitively requires the Frobenius method, the second family of solution can be obtained from the series expansion around the second irregular point $x=1$.

Change of variable : $t=x-1$ $$t(t+1)y''(t)+3(t+1)y'(t)+y(t)=0$$ $$y(t)=t^\mu\sum_{n=0}^\infty b_nt^n$$ Proceed as above, with $\mu=-2$

This will lead to $y(t)=t^{-2}\big((t+1)\ln(t+1)+1 \big)$.

Nevertheless, this method is boring compared to the reduction of order method.


With integer differences in integer roots, if you are lucky you get a polynomial solution. However in general there is interference so that you need to apply order reduction. Somewhere there is a theorem that exactly differentiates these cases.

So we are looking for $y_2(x)=y(x)=\frac{x}{(x-1)^2}u(x)$ with derivatives and differential equation for $u$ as follows.

\begin{align} y'(x)&=\left(-\frac1{(x-1)^2}-\frac2{(x-1)^3}\right)u(x)+\frac{x}{(x-1)^2}u'(x)\\ &=-\frac{x+1}{(x-1)^3}u(x)+\frac{x}{(x-1)^2}u'(x)\\ y''(x)&=\left(\frac2{(x-1)^3}+\frac6{(x-1)^4}\right)u(x)-2\frac{x+1}{(x-1)^3}u'(x)+\frac{x}{(x-1)^2}u''(x)\\ &=2\frac{x+2}{(x-1)^4}u(x)-2\frac{x+1}{(x-1)^3}u'(x)+\frac{x}{(x-1)^2}u''(x) \\[1em]\hline 0=x(x-1)y''(x)+3xy'(x)+y(x)&= \frac{x(x-2)}{(x-1)^2}u'(x)+\frac{x}{(x-1)^2}u''(x). \end{align} This leads to $((x-2)u'(x))'=0$ so that one non-trivial solution is $u(x)=\ln|1-x/2|$, giving the second basis solution as $$y_2(x)=\frac{x\ln|1-x/2|}{(x-1)^2}.$$ This actually should appear in the power series approach, starting with a free quadratic coefficient.

  • 1
    $\begingroup$ This is not the Frobenius method asked by the OP. $\endgroup$ – JJacquelin Mar 15 at 10:11

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