2.4.30 Problem 26

Internal problem ID [8919]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 26
Date solved : Wednesday, March 05, 2025 at 07:08:13 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

$$\begin{array}{rl}{x}^2(3+x)y^{\prime \prime }+5x(1+x)y^{\prime }-(1-4x)y& =0\end{array}$$

Using series expansion around $x=0$

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

$$(x^3+3x^2)y^{\prime \prime }+(5x^2+5x)y^{\prime }+(4x-1)y=0$$

The following is summary of singularities for the above ode. Writing the ode as

$$\begin{array}{rl}{y}^{\prime \prime }+p(x)y^{\prime }+q(x)y& =0\end{array}$$

Where

$$\begin{array}{rl}p(x)& =\frac{5+5x}{x(3+x)}\\ q(x)& =\frac{4x-1}{x^2(3+x)}\end{array}$$

Table 2.79: Table $p(x),q(x)$ singularites.

$p(x)=\frac{5+5x}{x(3+x)}$
singularity	type
$x=-3$	$\text{“regular”}$
$x=0$	$\text{“regular”}$

$q(x)=\frac{4x-1}{x^2(3+x)}$
singularity	type
$x=-3$	$\text{“regular”}$
$x=0$	$\text{“regular”}$

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : $[-3,0,\mathrm{\infty }]$

Irregular singular points : $[]$

Since $x=0$ is regular singular point, then Frobenius power series is used. The ode is normalized to be

$$x^2(3+x)y^{\prime \prime }+(5x^2+5x)y^{\prime }+(4x-1)y=0$$

Let the solution be represented as Frobenius power series of the form

$$y=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}$$

Then

$$\begin{array}{rl}{y}^{\prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above back into the ode gives

$$\begin{array}{}\text{(1)}& x^2(3+x)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})+(5x^2+5x)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1})+(4x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{1+n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{1+n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}4x^{1+n+r}{a}_n)+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-a_n{x}^{n+r})=0\end{array}$$

The next step is to make all powers of $x$ be $n+r$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n+r}$ and adjusting the power and the corresponding index gives

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{1+n+r}{a}_n(n+r)(n+r-1)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}(n+r-1)(n+r-2)x^{n+r}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{1+n+r}{a}_n(n+r)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}5a_{n-1}(n+r-1)x^{n+r}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}4x^{1+n+r}{a}_n& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}4a_{n-1}{x}^{n+r}\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n+r$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}(n+r-1)(n+r-2)x^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^{n+r}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}5a_{n-1}(n+r-1)x^{n+r})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}5x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}4a_{n-1}{x}^{n+r})+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-a_n{x}^{n+r})=0\end{array}$$

The indicial equation is obtained from $n=0$. From Eq (2B) this gives

$$3x^{n+r}{a}_n(n+r)(n+r-1)+5x^{n+r}{a}_n(n+r)-a_n{x}^{n+r}=0$$

When $n=0$ the above becomes

$$3x^r{a}_{0}r(-1+r)+5x^r{a}_{0}r-a_0{x}^r=0$$

Or

$$(3x^{r}r(-1+r)+5x^{r}r-x^r)a_0=0$$

Since $a_0\ne 0$ then the above simplifies to

$$(3r^2+2r-1)x^r=0$$

Since the above is true for all $x$ then the indicial equation becomes

$$3r^2+2r-1=0$$

Solving for $r$ gives the roots of the indicial equation as

$$\begin{array}{rl}{r}_1& =\frac{1}{3}\\ r_2& =-1\end{array}$$

Since $a_0\ne 0$ then the indicial equation becomes

$$(3r^2+2r-1)x^r=0$$

Solving for $r$ gives the roots of the indicial equation as $[\frac{1}{3},-1]$.

Since $r_1-r_2=\frac{4}{3}$ is not an integer, then we can construct two linearly independent solutions

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{r_1}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =x^{r_2}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+\frac{1}{3}}\\ y_2\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n-1}\end{array}$$

We start by finding $y_1\left(x\right)$. Eq (2B) derived above is now used to find all $a_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $a_0$ is arbitrary and taken as $a_0=1$. For $1\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& a_{n-1}(n+r-1)(n+r-2)+3a_n(n+r)(n+r-1)+5a_{n-1}(n+r-1)+5a_n(n+r)+4a_{n-1}-a_n=0\end{array}$$

Solving for $a_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& a_n=-\frac{(1+n+r)a_{n-1}}{3n+3r-1}\end{array}$$

Which for the root $r=\frac{1}{3}$ becomes

$$\begin{array}{}\text{(5)}& a_n=-\frac{(4+3n)a_{n-1}}{9n}\end{array}$$

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=\frac{1}{3}$ and after as more terms are found using the above recursive equation.

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$

For $n=1$, using the above recursive equation gives

$$a_1=\frac{-2-r}{2+3r}$$

Which for the root $r=\frac{1}{3}$ becomes

$$a_1=-\frac{7}{9}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{2+3r}$	$-\frac{7}{9}$

For $n=2$, using the above recursive equation gives

$$a_2=\frac{r^2+5r+6}{9r^2+21r+10}$$

Which for the root $r=\frac{1}{3}$ becomes

$$a_2=\frac{35}{81}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{2+3r}$	$-\frac{7}{9}$
$a_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$\frac{35}{81}$

For $n=3$, using the above recursive equation gives

$$a_3=\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$$

Which for the root $r=\frac{1}{3}$ becomes

$$a_3=-\frac{455}{2187}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{2+3r}$	$-\frac{7}{9}$
$a_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$\frac{35}{81}$
$a_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$-\frac{455}{2187}$

For $n=4$, using the above recursive equation gives

$$a_4=\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$$

Which for the root $r=\frac{1}{3}$ becomes

$$a_4=\frac{1820}{19683}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{2+3r}$	$-\frac{7}{9}$
$a_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$\frac{35}{81}$
$a_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$-\frac{455}{2187}$
$a_4$	$\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$	$\frac{1820}{19683}$

For $n=5$, using the above recursive equation gives

$$a_5=\frac{-r^5-20r^4-155r^3-580r^2-1044r-720}{243r^5+3240r^4+16065r^3+36360r^2+36492r+12320}$$

Which for the root $r=\frac{1}{3}$ becomes

$$a_5=-\frac{6916}{177147}$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{-2-r}{2+3r}$	$-\frac{7}{9}$
$a_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$\frac{35}{81}$
$a_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$-\frac{455}{2187}$
$a_4$	$\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$	$\frac{1820}{19683}$
$a_5$	$\frac{-r^5-20r^4-155r^3-580r^2-1044r-720}{243r^5+3240r^4+16065r^3+36360r^2+36492r+12320}$	$-\frac{6916}{177147}$

Using the above table, then the solution $y_1\left(x\right)$ is

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{1/3}(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x^{1/3}(1-\frac{7x}{9}+\frac{35x^2}{81}-\frac{455x^3}{2187}+\frac{1820x^4}{19683}-\frac{6916x^5}{177147}+O\left(x^6\right))\end{array}$$

Now the second solution $y_2\left(x\right)$ is found. Eq (2B) derived above is now used to find all $b_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $b_0$ is arbitrary and taken as $b_0=1$. For $1\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& b_{n-1}(n+r-1)(n+r-2)+3b_n(n+r)(n+r-1)+5b_{n-1}(n+r-1)+5b_n(n+r)+4b_{n-1}-b_n=0\end{array}$$

Solving for $b_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& b_n=-\frac{(1+n+r)b_{n-1}}{3n+3r-1}\end{array}$$

Which for the root $r=-1$ becomes

$$\begin{array}{}\text{(5)}& b_n=-\frac{n{b}_{n-1}}{3n-4}\end{array}$$

At this point, it is a good idea to keep track of $b_n$ in a table both before substituting $r=-1$ and after as more terms are found using the above recursive equation.

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$

For $n=1$, using the above recursive equation gives

$$b_1=\frac{-2-r}{2+3r}$$

Which for the root $r=-1$ becomes

$$b_1=1$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$\frac{-2-r}{2+3r}$	$1$

For $n=2$, using the above recursive equation gives

$$b_2=\frac{r^2+5r+6}{9r^2+21r+10}$$

Which for the root $r=-1$ becomes

$$b_2=-1$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$\frac{-2-r}{2+3r}$	$1$
$b_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$-1$

For $n=3$, using the above recursive equation gives

$$b_3=\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$$

Which for the root $r=-1$ becomes

$$b_3=\frac{3}{5}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$\frac{-2-r}{2+3r}$	$1$
$b_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$-1$
$b_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$\frac{3}{5}$

For $n=4$, using the above recursive equation gives

$$b_4=\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$$

Which for the root $r=-1$ becomes

$$b_4=-\frac{3}{10}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$\frac{-2-r}{2+3r}$	$1$
$b_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$-1$
$b_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$\frac{3}{5}$
$b_4$	$\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$	$-\frac{3}{10}$

For $n=5$, using the above recursive equation gives

$$b_5=\frac{-r^5-20r^4-155r^3-580r^2-1044r-720}{243r^5+3240r^4+16065r^3+36360r^2+36492r+12320}$$

Which for the root $r=-1$ becomes

$$b_5=\frac{3}{22}$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$\frac{-2-r}{2+3r}$	$1$
$b_2$	$\frac{r^2+5r+6}{9r^2+21r+10}$	$-1$
$b_3$	$\frac{-r^3-9r^2-26r-24}{27r^3+135r^2+198r+80}$	$\frac{3}{5}$
$b_4$	$\frac{r^4+14r^3+71r^2+154r+120}{81r^4+702r^3+2079r^2+2418r+880}$	$-\frac{3}{10}$
$b_5$	$\frac{-r^5-20r^4-155r^3-580r^2-1044r-720}{243r^5+3240r^4+16065r^3+36360r^2+36492r+12320}$	$\frac{3}{22}$

Using the above table, then the solution $y_2\left(x\right)$ is

$$\begin{array}{rl}{y}_2\left(x\right)& =x^{1/3}(b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6\dots )\\ & =\frac{1+x-x^2+\frac{3x^3}{5}-\frac{3x^4}{10}+\frac{3x^5}{22}+O\left(x^6\right)}{x}\end{array}$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1{x}^{1/3}(1-\frac{7x}{9}+\frac{35x^2}{81}-\frac{455x^3}{2187}+\frac{1820x^4}{19683}-\frac{6916x^5}{177147}+O\left(x^6\right))+\frac{c_2(1+x-x^2+\frac{3x^3}{5}-\frac{3x^4}{10}+\frac{3x^5}{22}+O\left(x^6\right))}{x}\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_1{x}^{1/3}(1-\frac{7x}{9}+\frac{35x^2}{81}-\frac{455x^3}{2187}+\frac{1820x^4}{19683}-\frac{6916x^5}{177147}+O\left(x^6\right))+\frac{c_2(1+x-x^2+\frac{3x^3}{5}-\frac{3x^4}{10}+\frac{3x^5}{22}+O\left(x^6\right))}{x}\end{array}$$

✓ Maple. Time used: 0.015 (sec). Leaf size: 48

Order:=6; 
ode:=x^2*(x+3)*diff(diff(y(x),x),x)+5*x*(x+1)*diff(y(x),x)-(1-4*x)*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=\frac{c_2{x}^{4/3}(1-\frac{7}{9}x+\frac{35}{81}{x}^2-\frac{455}{2187}{x}^3+\frac{1820}{19683}{x}^4-\frac{6916}{177147}{x}^5+\mathrm{O}\left(x^6\right))+c_1(1+x-x^2+\frac{3}{5}{x}^3-\frac{3}{10}{x}^4+\frac{3}{22}{x}^5+\mathrm{O}\left(x^6\right))}{x}$$

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
<- linear_1 successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2(x+3)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)+5x(x+1)\left(\frac{d}{dx}y\left(x\right)\right)-(1-4x)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=-\frac{(4x-1)y\left(x\right)}{x^2(x+3)}-\frac{5(x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x+3)}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)+\frac{5(x+1)\left(\frac{d}{dx}y\left(x\right)\right)}{x(x+3)}+\frac{(4x-1)y\left(x\right)}{x^2(x+3)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{5(x+1)}{x(x+3)},P_3\left(x\right)=\frac{4x-1}{x^2(x+3)}]\\ & \circ & (x+3)\cdot P_2\left(x\right)\text{is analytic at}x=-3\\ & & ((x+3)\cdot P_2\left(x\right))|{}_{x=-3}=\frac{10}{3}\\ & \circ & {(x+3)}^2\cdot P_3\left(x\right)\text{is analytic at}x=-3\\ & & ({(x+3)}^2\cdot P_3\left(x\right))|{}_{x=-3}=0\\ & \circ & x=-3\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=-3\\ \bullet & & \text{Multiply by denominators}\\ & & x^2(x+3)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)+5x(x+1)\left(\frac{d}{dx}y\left(x\right)\right)+(4x-1)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-3\text{so that the regular singular point is at}u=0\\ & & (u^3-6u^2+9u)\left(\frac{d^2}{d{u}^2}y\left(u\right)\right)+(5u^2-25u+30)\left(\frac{d}{du}y\left(u\right)\right)+(4u-13)y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot y\left(u\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{u}^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=0..2\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)\text{to series expansion for}m=1..3\\ & & u^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & 3a_{0}r(7+3r)u^{-1+r}+(3a_1(1+r)(10+3r)-a_0(13+6r)(1+r))u^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(3a_{k+1}(k+r+1)(3k+10+3r)-a_k(6k+6r+13)(k+r+1)+a_{k-1}{(k+r+1)}^2)u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 3r(7+3r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,-\frac{7}{3}\}\\ \bullet & & \text{Each term must be 0}\\ & & 3a_1(1+r)(10+3r)-a_0(13+6r)(1+r)=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -6((a_k-\frac{a_{k-1}}{6}-\frac{3a_{k+1}}{2})k+(a_k-\frac{a_{k-1}}{6}-\frac{3a_{k+1}}{2})r+\frac{13a_k}{6}-\frac{a_{k-1}}{6}-5a_{k+1})(k+r+1)=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & -6((a_{k+1}-\frac{a_k}{6}-\frac{3a_{k+2}}{2})(k+1)+(a_{k+1}-\frac{a_k}{6}-\frac{3a_{k+2}}{2})r+\frac{13a_{k+1}}{6}-\frac{a_k}{6}-5a_{k+2})(k+r+2)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}+r{a}_k-6r{a}_{k+1}+2a_k-19a_{k+1}}{3(3k+13+3r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}+2a_k-19a_{k+1}}{3(3k+13)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}+2a_k-19a_{k+1}}{3(3k+13)},30a_1-13a_0=0]\\ \bullet & & \text{Revert the change of variables}u=x+3\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+3)}^k,a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}+2a_k-19a_{k+1}}{3(3k+13)},30a_1-13a_0=0]\\ \bullet & & \text{Recursion relation for}r=-\frac{7}{3}\\ & & a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}-\frac{1}{3}{a}_k-5a_{k+1}}{3(3k+6)}\\ \bullet & & \text{Solution for}r=-\frac{7}{3}\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k-\frac{7}{3}},a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}-\frac{1}{3}{a}_k-5a_{k+1}}{3(3k+6)},-12a_1-\frac{4a_0}{3}=0]\\ \bullet & & \text{Revert the change of variables}u=x+3\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+3)}^{k-\frac{7}{3}},a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}-\frac{1}{3}{a}_k-5a_{k+1}}{3(3k+6)},-12a_1-\frac{4a_0}{3}=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+3)}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(x+3)}^{k-\frac{7}{3}}),a_{k+2}=-\frac{k{a}_k-6k{a}_{k+1}+2a_k-19a_{k+1}}{3(3k+13)},30a_1-13a_0=0,b_{k+2}=-\frac{k{b}_k-6k{b}_{k+1}-\frac{1}{3}{b}_k-5b_{k+1}}{3(3k+6)},-12b_1-\frac{4b_0}{3}=0]\end{array}$$

✓ Mathematica. Time used: 0.016 (sec). Leaf size: 82

ode=x^2*(3+x)*D[y[x],{x,2}] + 5*x*(1+x)*D[y[x],x] - (1-4*x)*y[x] == 0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1\sqrt[3]{x}(-\frac{6916x^5}{177147}+\frac{1820x^4}{19683}-\frac{455x^3}{2187}+\frac{35x^2}{81}-\frac{7x}{9}+1)+\frac{c_2(\frac{3x^5}{22}-\frac{3x^4}{10}+\frac{3x^3}{5}-x^2+x+1)}{x}$$

✓ Sympy. Time used: 1.118 (sec). Leaf size: 15

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(x + 3)*Derivative(y(x), (x, 2)) + 5*x*(x + 1)*Derivative(y(x), x) - (1 - 4*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

$$y\left(x\right)=C_2\sqrt[3]{x}+\frac{C_1}{x}+O\left(x^6\right)$$