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$2x^2y''-xy'+(x-5)y=0$

I know how to solve using power series but I am not able to understand the Frobenius method. Can somebody do the solution in a basic manner so that I can understand how to solve this about point x=0

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  • $\begingroup$ You mean in $x=0$? First solve the Euler-Cauchy equation $2x^2y_0''-xy_0'-5y_0=0$ and then find the basis solutions of the given equation as perturbations of it, $y(x)=y_0(x)u(x)$. $\endgroup$ – Dr. Lutz Lehmann Dec 2 at 23:26
  • $\begingroup$ What is the point about which the solution is asked. Is it 0 ?? $\endgroup$ – Om Prakash Dec 3 at 18:04
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$$2x^2y''-xy'+(x-5)y=0\tag1$$ Comparing the given equation with equation $(A1)$, we have $~P(x)=-\dfrac{1}{2x}~$and $~Q(x)=\dfrac{(x-5)}{2x^2}~.$

Here $~x=0~$ is a singular point of the given differential equation.

Now $$\lim_{x\to 0}~(x-0)P(x)=-\dfrac{1}{2}~\lim_{x\to 0}~(1)=-\dfrac{1}{2}$$ $$\lim_{x\to 0}~(x-0)^2 Q(x)=\dfrac{1}{2}~\lim_{x\to 0}~(x-5)=-\dfrac{5}{2}$$

So $~x=0~$ is a regular singular point and then there exists at least one solution of the form $$y = x^r \sum_{k=0}^\infty a_k x^k =\sum_{k=0}^\infty a_k x^{k+r}~\qquad \text{where$~~a_0 \ne 0$}~.\tag2$$The constant $~r~$and the coefficient$~a_k$'s are to be determined.

Now $$y'=\sum_{k=0}^\infty a_k(k+r) x^{k+r-1}\tag3$$ $$y''=\sum_{k=0}^\infty a_k(k+r)(k+r-1) x^{k+r-2}\tag4$$ Using equation$(2),~(3)~$and$~(4)~$, from $(1)$ we have $$2x^2\sum_{k=0}^\infty a_k(k+r)(k+r-1) x^{k+r-2}-x\sum_{k=0}^\infty a_k(k+r) x^{k+r-1}+(x-5)\sum_{k=0}^\infty a_k x^{k+r}=0$$ $$\implies 2\sum_{k=0}^\infty a_k(k+r)(k+r-1) x^{k+r}-\sum_{k=0}^\infty a_k(k+r) x^{k+r}+\sum_{k=0}^\infty a_k x^{k+r+1}-5\sum_{k=0}^\infty a_k x^{k+r}=0$$ $$\implies \sum_{k=0}^\infty a_k\{2(k+r)(k+r-1)-(k+r)-5\}x^{k+r}+\sum_{k=0}^\infty a_k x^{k+r+1}=0\tag5$$ The coefficient of each power of$~x~$appearing in the left hand side must vanish. We observe that the first term$~(k = 0)~$of the first sum is a term with$~x^r~$, while the first term$~(k = 0)~$of the second sum is a term with$~x^{r+1}~$. In other words, the coefficient of$~x^r~$in the entire left hand side of the last equation is $$a_0\{2(0+r)(0+r-1)-(0+r)-5\}=[2r^2-3r-5]a_0$$ So the indicial equation is $$(2r^2-3r-5)a_0=0\implies 2r^2-3r-5=0$$as $~a_0\ne 0~$.

Roots of the indicial equation are $~r=\dfrac{5}{2},~-1~$.

Since difference between two exponents is not a positive integer, two independent Frobenius series corresponding to $~r=-1~$and$~r=\dfrac{5}{2}~$exist.

In order to determine the series we have the following recurrence formula as derived from equation $(5)$ : $$ a_{k+1}\{2(k+r+1)(k+r)-(k+r+1)-5\}+a_k =0$$ $$\implies a_{k+1}=-\dfrac{1}{(k+r+1)(2k+2r-1)-5}a_k$$ $$\implies a_{k+1}=-\dfrac{1}{(k+r+2)(2k+2r-3)}a_k\tag6$$ For$~r=-1~$, we have from $(6)$, $$a_{k+1}=-\dfrac{1}{(k+1)(2k-5)}a_k$$ So $~a_1=\dfrac{1}{5}a_0~$

$~a_2=\dfrac{1}{6}a_1=\dfrac{1}{6}\cdot \dfrac{1}{5}a_0~$

$~a_3=\dfrac{1}{3}a_2=\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot \dfrac{1}{5}a_0~$ and so on.

For$~r=\dfrac{5}{2}~$, we have from $(6)$, $$a_{k+1}=-\dfrac{1}{(2k+9)(k+1)}a_k$$ So $~a_1=-\dfrac{1}{9}a_0~$

$~a_2=-\dfrac{1}{22}a_1=\dfrac{1}{22}\cdot \dfrac{1}{9}a_0~$

$~a_3=-\dfrac{1}{39}a_2=-\dfrac{1}{39}\cdot\dfrac{1}{22}\cdot \dfrac{1}{9}a_0~$ and so on.

Thus the two independent Frobenius series solution corresponding to $~r=-1~$and$~r=\dfrac{5}{2}~$are $$y_1=a_0~x^{-1}\left(1+\dfrac{1}{5}x+\dfrac{1}{6}\cdot \dfrac{1}{5}x^2+\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot \dfrac{1}{5}x^3+\cdots\right)$$ $$y_2=a_0~x^{{5}/{2}}\left(1-\dfrac{1}{9}x+\dfrac{1}{22}\cdot \dfrac{1}{9}x^2-\dfrac{1}{39}\cdot\dfrac{1}{22}\cdot \dfrac{1}{9}x^3+\cdots\right)$$ The general solution of the equation $(1)$ is $$y=cy_1+dy_2$$ $$=a_0\left[c~x^{-1}\left(1+\dfrac{1}{5}x+\dfrac{1}{6}\cdot \dfrac{1}{5}x^2+\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot \dfrac{1}{5}x^3+\cdots\right)+d~x^{{5}/{2}}\left(1-\dfrac{1}{9}x+\dfrac{1}{22}\cdot \dfrac{1}{9}x^2-\dfrac{1}{39}\cdot\dfrac{1}{22}\cdot \dfrac{1}{9}x^3+\cdots\right)\right]$$where $~c,~d~$are constants.

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Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0\tag{A1}$$ where $x \in D \subseteq \mathbb{C}$.

The point $x_0 \in D$ is said to be an ordinary point of the above given differential equation if $P(x)$ and $Q(x)$ are analytic at $x_0$.

If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$, the point $x_0$ is called a singular point of the given differential equation.

A singular point $x_0$ of the given differential equation is said to be regular singular point if the function $(x-x_0)P(x)$ and $(x-x_0)^2 Q(x)$ are analytic at $x_0$ and irregular otherwise.


Method of Frobenius

Suppose that $$𝑝(𝑥)𝑦″+𝑞(𝑥)𝑦′+𝑟(𝑥)𝑦=0$$has a regular singular point at  $~𝑥=0~$ , then there exists at least one solution of the form $$𝑦=𝑥^𝑟\sum_{𝑘=0}^∞ 𝑎_𝑘𝑥^𝑘~.$$ A solution of this form is called a Frobenius-type solution.


Indicial equation (Direct Method): If $x=\alpha$ is a regular singular point of the given differential equation $$u''+P(x)u'+Q(x)u=0$$ then the indicial equation is $$r(r-1)+p_0r+q_0=0$$ where

$$p_0=\lim_{x\to \alpha }(x-\alpha)P(x)$$

$$q_0=\lim_{x\to \alpha} (x-\alpha)^2Q(x)$$

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  • $\begingroup$ In the recursion the denominator $(a+1)(2a-1)-5=2a^2+a-6$ can be factorized as $(a+2)(2a-3)$, $a=k+r$. $\endgroup$ – Dr. Lutz Lehmann Dec 5 at 20:11
  • $\begingroup$ Can you tell me how did u conclude x=0 is noy analytic at x=0?What are the conditions to check $\endgroup$ – Aladdin Dec 6 at 3:14
  • $\begingroup$ Since difference between two exponents is not a positive integer, two independent Frobenius series corresponding to r=−1 and r=5/2 exist. But the difference between r2 and r1 is positive? $\endgroup$ – Aladdin Dec 6 at 3:16
  • $\begingroup$ Please mentioned where I conclude " $x=0$ is not analytic at $x=0$" @Aladdin $\endgroup$ – nmasanta Dec 6 at 3:46
  • $\begingroup$ In connection to your second comment @Aladdin, I would like to mentioned that the although the difference between $r_1$ and $r_2$ is positive yet it is not an integer (you skip this point). $\endgroup$ – nmasanta Dec 6 at 3:51

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