Problem: differential equation Hi I try solve the following problem of differential equation
$$ x''+tx'+\frac{1}{1+t+t^2}x=0\tag 1$$ when $$x(1)=0\ \ \ ;\ \ \ x'(1)=1 $$
is the solution analytic in $t_0=1$ and his convergence radius is $R>1$?
Ok, I think I need put the differential equation like a frobenius differential equation, then I get, with the initial equation, that my solution is define by $$ \varphi_1(t)=\sum_{n=0}^{\infty} a_n(t-1)^{n+1}$$
$$\varphi_2(t)=C\varphi_1(t)\log(t-1)+\sum_{n=0}^{\infty} b_n(t-1)^n $$
I can not work with $\varphi_1$ in $(1)$, I do not know... someone could help me?
 A: It's pretty easy to show by substitution that $t=1$ is a regular point for this equation, since all the coefficients have finite values at this point.
Which means, we can search for solution as a regular power series around $t=1$, and the Frobenius method is not needed here.
It makes more sence to introduce a new variable:
$$u=t-1, \qquad y(u)=x(t)$$
Then our equation becomes:
$$y''+(1+u)y'+\frac{1}{3+3u+u^2}y=0$$
As the denominator is not $0$ for $u \to 0$, we can multiply the equation by it:
$$(3+3u+u^2)y''+(3+6u+4u^2+u^3) y'+y=0$$
Now we search for the solution in the form:
$$y=\sum_{n=0}^\infty a_n u^n$$
Substitution gives us the following recurrence:
$$3n(n-1)a_n+3(n-1)^2 a_{n-1} +(n^2+n-5) a_{n-2}+4(n-3)a_{n-3}+(n-4)a_{n-4}=0 \tag{2}$$
$$n \geq 4$$
From initial conditions and appropriate equations for initial terms we can see that:
$$a_0=0$$
$$a_1=1$$
$$a_2=-\frac12$$
$$a_3=-\frac{1}{18}$$
Now we can use the recurrence (2) to get every other coefficient:
$$a_4=\frac{5}{36}$$
$$\cdots$$

We can go back to $x(t)$ by writing:
$$x=\sum_{n=0}^\infty a_n (t-1)^n$$
