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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?

Sam
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1 Answers1

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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:

$$x=\sum_{n=0}^\infty a_n (t-1)^n$$

Yuriy S
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