2.4.58 Problem 55

Internal problem ID [8946]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 55
Date solved : Friday, April 25, 2025 at 05:25:30 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

$$\begin{array}{rl}2x^2{y}^{\prime \prime }+3x{y}^{\prime }-y& =0\end{array}$$

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.

$$2x^2{y}^{\prime \prime }+3x{y}^{\prime }-y=0$$

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

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

Table 2.109: Table $p(x),q(x)$ singularites.

$p(x)=\frac{3}{2x}$
singularity	type
$x=0$	$\text{“regular”}$

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

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

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

Irregular singular points : $[]$

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

$$2x^2{y}^{\prime \prime }+3x{y}^{\prime }-y=0$$

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

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

Which simplifies to

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

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

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

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

$$2x^{n+r}{a}_n(n+r)(n+r-1)+3x^{n+r}{a}_n(n+r)-a_n{x}^{n+r}=0$$

When $n=0$ the above becomes

$$2x^r{a}_{0}r(-1+r)+3x^r{a}_{0}r-a_0{x}^r=0$$

Or

$$(2x^{r}r(-1+r)+3x^{r}r-x^r)a_0=0$$

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

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

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

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

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

$$\begin{array}{rl}{r}_1& =\frac{1}{2}\\ r_2& =-1\end{array}$$

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

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

Solving for $r$ gives the roots of the indicial equation as $[\frac{1}{2},-1]$.

Since $r_1-r_2=\frac{3}{2}$ 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

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+\frac{1}{2}}\\ y_2\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n-1}\end{array}$$

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 $0\le n$ the recursive equation is

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

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

$$\begin{array}{}\text{(4)}& a_n=0\end{array}$$

Which for the root $r=\frac{1}{2}$ becomes

$$\begin{array}{}\text{(5)}& a_n=0\end{array}$$

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=\frac{1}{2}$ 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

$$a_1=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$

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

$$a_2=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$

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

$$a_3=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$

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

$$a_4=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$
$a_4$	$0$	$0$

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

$$a_5=0$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$0$	$0$
$a_2$	$0$	$0$
$a_3$	$0$	$0$
$a_4$	$0$	$0$
$a_5$	$0$	$0$

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

$$\begin{array}{rl}{y}_1\left(x\right)& =\sqrt{x}(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =\sqrt{x}(1+O\left(x^6\right))\end{array}$$

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 $0\le n$ the recursive equation is

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

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

$$\begin{array}{}\text{(4)}& b_n=0\end{array}$$

Which for the root $r=-1$ becomes

$$\begin{array}{}\text{(5)}& b_n=0\end{array}$$

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

$$b_1=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$

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

$$b_2=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$0$	$0$

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

$$b_3=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$0$	$0$
$b_3$	$0$	$0$

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

$$b_4=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$0$	$0$
$b_3$	$0$	$0$
$b_4$	$0$	$0$

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

$$b_5=0$$

And the table now becomes

$n$	$b_{n,r}$	$b_n$
$b_0$	$1$	$1$
$b_1$	$0$	$0$
$b_2$	$0$	$0$
$b_3$	$0$	$0$
$b_4$	$0$	$0$
$b_5$	$0$	$0$

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

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

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_1\sqrt{x}(1+O\left(x^6\right))+\frac{c_2(1+O\left(x^6\right))}{x}\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_1\sqrt{x}(1+O\left(x^6\right))+\frac{c_2(1+O\left(x^6\right))}{x}\end{array}$$

✓ Maple. Time used: 0.032 (sec). Leaf size: 20

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

$$y=\frac{x^{3/2}{c}_2+c_1}{x}+O\left(x^6\right)$$

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 
<- LODE of Euler type successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)-y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{y\left(x\right)}{2x^2}-\frac{3\left(\frac{d}{dx}y\left(x\right)\right)}{2x}\\ \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}{dx}\frac{d}{dx}y\left(x\right)+\frac{3\left(\frac{d}{dx}y\left(x\right)\right)}{2x}-\frac{y\left(x\right)}{2x^2}=0\\ \bullet & & \text{Multiply by denominators of the ODE}\\ & & 2x^2\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)-y\left(x\right)=0\\ \bullet & & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ ◻& & \text{Substitute the change of variables back into the ODE}\\ & \circ & \text{Calculate the}\text{1st}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}y\left(t\right)\right)\left(\frac{d}{dx}t\left(x\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}y\left(t\right)}{x}\\ & \circ & \text{Calculate the}\text{2nd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right){\left(\frac{d}{dx}t\left(x\right)\right)}^2+\left(\frac{d}{dx}\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dt}y\left(t\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & 2x^2(\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2})+3\frac{d}{dt}y\left(t\right)-y\left(t\right)=0\\ \bullet & & \text{Simplify}\\ & & 2\frac{d}{dt}\frac{d}{dt}y\left(t\right)+\frac{d}{dt}y\left(t\right)-y\left(t\right)=0\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=-\frac{\frac{d}{dt}y\left(t\right)}{2}+\frac{y\left(t\right)}{2}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)+\frac{\frac{d}{dt}y\left(t\right)}{2}-\frac{y\left(t\right)}{2}=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2+\frac{1}{2}r-\frac{1}{2}=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & \frac{(r+1)(2r-1)}{2}=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-1,\frac{1}{2})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(t\right)={\mathrm{e}}^{-t}\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(t\right)={\mathrm{e}}^{\frac{t}{2}}\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-t}+\mathit{C}\mathit{2}{\mathrm{e}}^{\frac{t}{2}}\\ \bullet & & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}}{x}+\mathit{C}\mathit{2}\sqrt{x}\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=\frac{\mathit{C}\mathit{1}}{x}+\mathit{C}\mathit{2}\sqrt{x}\end{array}$$

✓ Mathematica. Time used: 0.004 (sec). Leaf size: 18

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

$$y(x)\to c_1\sqrt{x}+\frac{c_2}{x}$$

✓ Sympy. Time used: 0.746 (sec). Leaf size: 15

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

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