# Determine the nature of the solution at $x=\infty$ of the differential equation $(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+12y=0.$

Determine the nature of the solution at $$x=\infty$$ of the differential equation $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+12y=0.$$ Find series solution at infinity of the above equation (if it exists).

This seemed a strange question to me. I recognized the equation given is nothing but a Legendre's differential equation. Also, it's a known fact, that the general solution of a Legendre's Equation, i.e $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y$$ is, $$y=c_0(1-\frac{n(n+1)}{2}x^2-\frac{c_0n(n+1)(6-n(n+1))}{24}x^4+\cdots)+c_1(x+\frac{(2-n(n+1))}{6}x^3 +\cdots)\tag 1,$$ where $$c_0$$ and $$c_1$$ are arbitrary constants.

In the given equation, $$n=3,$$ and we can just find the general solution by plugging $$n=3$$ in $$(1).$$

But I don't understand what is the value of $$y$$ at $$\infty$$ other than that, $$y\to\infty$$ if that's what they mean in the question.

Next, if I understand it correctly, then "series solution at infinity" makes no sense whatsoever. This is because, $$\infty$$ is not a real point, we can find series solution, about any real point, which may be an ordinary point or a regular singular point, but what is it they mean by this phrase above ?

• the $= 0$ part is missing from the differential equation? Commented Jun 30, 2023 at 15:16
• @KenHung Fixed it. Thanks for pointing it out. But what do you think about this in general ? Commented Jun 30, 2023 at 15:17
• @Gonçalo oh, have n't thought about doing it. Let me try 'gain. Commented Jun 30, 2023 at 16:55
• @Gonçalo I have added an answer. Is this what you meant? Commented Jul 1, 2023 at 6:33

This solution is inspired from the comment of @Gonçalo.

Indeed, $$\infty$$ is not a real number, so, we can't really, find the series solution at $$\infty$$ assuming it to be a singular point or an ordinary point, whatsoever. However, the question is not absurd, for we can check the power series solution at $$\infty$$, by simply noting the fact, that $$x\to \frac 10\implies x\to \infty.$$

We can accomplish this, by a change of variables, i.e let us assume, $$x=\frac 1t.$$ Then, examining the solution at $$t=0$$ will be equivalent to examining solutions at $$x=\infty.$$ We note that, power series solution around $$t=0$$ is possible, because $$0$$ is a real number unlike the notion of $$\infty.$$

So, as for the solution, here it is:

Given $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+12y=0.$$

Let $$x=\frac 1t.$$ This means, $$t=\frac 1x.$$

We have, $$\frac {dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=-\frac{1}{x^2}=-t^2\frac{dy}{dt},$$ and $$\frac{d^2y}{dx^2}=\frac{dy}{dt}-\frac{1}{x^2}\frac{d^2y}{dt^2}(-\frac{1}{x^2})=2t^3\frac{dy}{dt}+t^4\frac{d^2y}{dt^2}.$$

Substituting these values in the given equation, we find, $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+12y=0\implies \frac{(t^2-1)}{t^2}\big [2t^3\frac{dy}{dt}+t^4\frac{d^2y}{dt^2}\big ] -\frac 2t\big [-t^2\frac{dy}{dt}\big ]+12y=0\implies t^2(t^2-1)\frac{d^2y}{dt^2}+2t^3\frac{dy}{dt}+12y=0\tag 1$$

We note that $$t=0$$ is a regular singular point.

We use the method of Frobenius to find a power series solution, about $$t=0.$$

Let $$y=\sum_{n=0}^{\infty}c_nt^{n+r},$$ such that $$c_0\neq 0.$$

So we have, $$y'=\sum_{n=0}^{\infty}c_n(n+r)t^{n+r-1},$$ and $$y''=\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)t^{n+r-2}.$$

Substituting these values so obtained, in $$(1)$$ we have, $$t^2(t^2-1)\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)t^{n+r-2}+2t^3\sum_{n=0}^{\infty}c_n(n+r)t^{n+r-1}+12\sum_{n=0}^{\infty}c_nt^{n+r}=0\implies \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)t^{n+r+2}-\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)t^{n+r}+2\sum_{n=0}^{\infty}c_n(n+r)t^{n+r+2}+12\sum_{n=0}^{\infty}c_nt^{n+r}=0\implies \sum_{n=2}^{\infty}c_{n-2}(n+r-2)(n+r-3)c_{n-2}t^{n+r}-\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)t^{n+r}+2\sum_{n=2}^{\infty}c_{n-2}(n+r-2)t^{n+r}+12\sum_{n=0}^{\infty}c_nt^{n+r}=0\implies \sum_{n=2}^{\infty}\big [ c_{n-2}(n+r-2)(n+r-3)c_{n-2}-c_n(n+r)(n+r-1)+2c_{n-2}(n+r-2)+12c_n\big ]t^{n+r}-r(r-1)c_0t^r-(r+1)(r)c_1t^{r+1}+12c_0t^r+12c_1t^{r+1}=0\tag 2$$

Using $$(2)$$ we find that, $$(12-r^2+r)c_0=0\implies r^2-r-12=0\implies (r-4)(r+3)=0.$$

Let $$r=r_1=4$$ and $$r=r_2=-3.$$

From $$(2)$$ , we also have, the recurrence relation, $$c_{n-2}(n+r-2)(n+r-3)c_{n-2}-c_n(n+r)(n+r-1)+2c_{n-2}(n+r-2)+12c_n=0\tag 3$$

If $$r=4$$ then $$(3)$$ becomes $$c_n=\frac{(n+2)(n+3)}{n(n+7)}c_{n-2},\forall n\geq 2.$$

Further, we observe, $$c_1(12-r^2-r)=0\implies c_1=0$$ for both $$r=r_1,r_2.$$

• If $$n=2$$ then, $$c_2=\frac{10}{9}c_0$$

• If $$n=3$$ then, $$c_3=0$$ as $$c_1=0$$ and consequently all odd coefficients are zero.

• If $$n=4$$ then, $$c_4=\frac{35}{33}c_0,$$ and so on.

Using these relations $$y=y_1=t^4[c_0+c_1x+\cdots ]=t^4c_0\big[ 1+\frac {10}{9}t^2+\frac{35}{33}t^4+\cdots\big ]=\frac{1}{x^4}c_0\big[ 1+\frac {10}{9}\frac{1}{x^2}+\frac{35}{33}\frac{1}{x^4}+\cdots\big ].$$

Now, if $$r=r_2=-3,$$ then from $$(3)$$ we have, $$c_{n}=\frac{(n-5)(n-4)c_{n-2}}{n(n-7)},\forall n\geq 2.$$

Using these, we obtain $$c_2=-\frac 35c_0$$ and $$c_3=0$$ (as $$c_1=0$$) due to which all coefficients are equal to zero.

And $$c_4=0,$$ which means all the coefficients $$c_n=0$$ for all $$n\geq 4.$$

Using these relations, we find, $$y=y_2=t^{-3}(c_0+c_2t^2)=t^{-3}c_0(1-\frac 35t^2)=x^3c_0(1-\frac 35\frac{1}{x^2}).$$

Hence, the general solution of the given differential equation is, $$y=c_1y_1(x)+c_2y_2(x)=c_1\frac{1}{x^4}\big[ 1+\frac {10}{9}\frac{1}{x^2}+\frac{35}{33}\frac{1}{x^4}+\cdots\big ]+c_2x^3(1-\frac 35\frac{1}{x^2}).$$

This is the nature of the solution at $$x=\infty.$$

Let $$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0$$ be Legendre's differential equation.

Then the general solution is $$y(x)=k_1\cdot Q_n(x)+k_2\cdot P_n(x)$$

with $$P_n(x), Q_n(x)$$ the Legendre polynomials.

For the special case $$n=3$$ we can find the real valued solution for $$x>1$$:

$$y(x)=c_{1}\cdot \left(\left(\frac{1}{8} x-\frac{5}{24} x^{3}\right) \left(\ln\left(x+1\right)-\ln\left(x-1\right)\right)-\frac{1}{9}+\frac{5}{12} x^{2}\right)+c_{2} \left(x-\frac{5}{3} x^{3}\right)$$

Note that $$P_3(x)=\frac{1}{2} \left(5 x^3-3 x\right)$$

Now we do the series expansion at $$x\to \infty$$ and obtain:

$$y(x)|_{x=\infty}\simeq c_{2}\cdot(x-\frac{5}{3} x^{3} )-c_{1}\cdot(\frac{1}{105 x^{4}}+\frac{2}{189 x^{6}}+\frac{1}{99 x^{8}})+\mathrm{O}\! \left(\frac{1}{x^{9}}\right)$$