2.5.2 Problem 2

Internal problem ID [8962]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 2
Date solved : Friday, April 25, 2025 at 05:26:58 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order solved by an integrating factor

Time used: 0.070 (sec)

Solve

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

The ode satisfies this form

$$y^{\prime \prime }+p\left(x\right)y^{\prime }+\frac{(p^{\prime }\left(x\right)+p{\left(x\right)}^2)y}{2}=f\left(x\right)$$

Where $p(x)=2x$. Therefore, there is an integrating factor given by

$$\begin{array}{rl}M(x)& =e^{\frac{1}{2}\int pdx}\\ & =e^{\int 2xdx}\\ & ={\mathrm{e}}^{\frac{x^2}{2}}\end{array}$$

Multiplying both sides of the ODE by the integrating factor $M(x)$ makes the left side of the ODE a complete differential

$$\begin{array}{rl}{(M(x)y)}^{″}& =0\\ {\left({\mathrm{e}}^{\frac{x^2}{2}}y\right)}^{″}& =0\end{array}$$

Integrating once gives

$${\left({\mathrm{e}}^{\frac{x^2}{2}}y\right)}^{\prime }=c_1$$

Integrating again gives

$$\left({\mathrm{e}}^{\frac{x^2}{2}}y\right)=c_{1}x+c_2$$

Hence the solution is

$$\begin{array}{rl}y& =\frac{c_{1}x+c_2}{{\mathrm{e}}^{\frac{x^2}{2}}}\end{array}$$

Or

$$y=c_{1}x{\mathrm{e}}^{-\frac{x^2}{2}}+c_2{\mathrm{e}}^{-\frac{x^2}{2}}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_{1}x{\mathrm{e}}^{-\frac{x^2}{2}}+c_2{\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Solved as second order ode using change of variable on y method 1

Time used: 0.092 (sec)

Solve

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

In normal form the given ode is written as

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =2x\\ q\left(x\right)& =x^2+1\end{array}$$

Calculating the Liouville ode invariant $Q$ given by

$$\begin{array}{rl}Q& =q-\frac{p^{\prime }}{2}-\frac{p^2}{4}\\ & =x^2+1-\frac{{\left(2x\right)}^{\prime }}{2}-\frac{{\left(2x\right)}^2}{4}\\ & =x^2+1-\frac{\left(2\right)}{2}-\frac{\left(4x^2\right)}{4}\\ & =x^2+1-\left(1\right)-x^2\\ & =0\end{array}$$

Since the Liouville ode invariant does not depend on the independent variable $x$ then the transformation

$$\begin{array}{}\text{(3)}& y=v\left(x\right)z\left(x\right)\end{array}$$

is used to change the original ode to a constant coefficients ode in $v$. In (3) the term $z\left(x\right)$ is given by

$$\begin{array}{rl}z\left(x\right)& ={\mathrm{e}}^{-\int \frac{p\left(x\right)}{2}dx}\\ & =e^{-\int \frac{2x}{2}}\\ \text{(5)}& & ={\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Hence (3) becomes

$$\begin{array}{}\text{(4)}& y=v\left(x\right){\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Applying this change of variable to the original ode results in

$$\begin{array}{r}{v}^{\prime \prime }\left(x\right){\mathrm{e}}^{-\frac{x^2}{2}}=0\end{array}$$

Which is now solved for $v\left(x\right)$.

The above ode can be simplified to

$$\begin{array}{r}{v}^{\prime \prime }\left(x\right)=0\end{array}$$

Integrating twice gives the solution

$$v=c_{1}x+c_2$$

Will add steps showing solving for IC soon.

Now that $v$ is known, then

$$\begin{array}{rl}y\left(x\right)& =vz\left(x\right)\\ \text{(7)}& & =(c_{1}x+c_2)\left(z\left(x\right)\right)\end{array}$$

But from (5)

$$\begin{array}{rl}z\left(x\right)& ={\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Hence (7) becomes

$$\begin{array}{r}y\left(x\right)=(c_{1}x+c_2){\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =(c_{1}x+c_2){\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.145 (sec)

Solve

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

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+2x{y}^{\prime }+(x^2+1)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =2xC& =x^2+1\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =0\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.115: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0--\mathrm{\infty }\\ & =\mathrm{\infty }\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $infinity$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=0$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)=1$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{2x}{1}dx}\\ & =z_1{e}^{-\frac{x^2}{2}}\\ & =z_1\left({\mathrm{e}}^{-\frac{x^2}{2}}\right)\end{array}$$

Which simplifies to

$$y_1={\mathrm{e}}^{-\frac{x^2}{2}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{2x}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-x^2}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(x\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{-\frac{x^2}{2}}\right)+c_2\left({\mathrm{e}}^{-\frac{x^2}{2}}\left(x\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{-\frac{x^2}{2}}+c_{2}x{\mathrm{e}}^{-\frac{x^2}{2}}\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 16

ode:=diff(diff(y(x),x),x)+2*x*diff(y(x),x)+(x^2+1)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y={\mathrm{e}}^{-\frac{x^2}{2}}(c_{2}x+c_1)$$

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)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+2x\left(\frac{d}{dx}y\left(x\right)\right)+(x^2+1)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{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{x}^m\cdot y\left(x\right)\text{to series expansion for}m=0..2\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)}}{a}_k{x}^{k+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)+m}}{a}_{k-m}{x}^k\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k}k{x}^k\\ & \circ & \text{Convert}\frac{d}{dx}\frac{d}{dx}y\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k}k(k-1)x^{k-2}\\ & \circ & \text{Shift index using}k->k+2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)x^k\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_2+a_0+(6a_3+3a_1)x+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_{k+2}(k+2)(k+1)+a_k(2k+1)+a_{k-2})x^k)=0\\ \bullet & & \text{The coefficients of each power of}x\text{must be 0}\\ & & [2a_2+a_0=0,6a_3+3a_1=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_2=-\frac{a_0}{2},a_3=-\frac{a_1}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k^2+3k+2)a_{k+2}+2a_{k}k+a_k+a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & ({(k+2)}^2+3k+8)a_{k+4}+2a_{k+2}(k+2)+a_{k+2}+a_k=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+4}=-\frac{2k{a}_{k+2}+a_k+5a_{k+2}}{k^2+7k+12},a_2=-\frac{a_0}{2},a_3=-\frac{a_1}{2}]\end{array}$$

✓ Mathematica. Time used: 0.024 (sec). Leaf size: 22

ode=D[y[x],{x,2}]+2*x*D[y[x],x]+(x^2+1)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to e^{-\frac{x^2}{2}}(c_{2}x+c_1)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(2*x*Derivative(y(x), x) + (x**2 + 1)*y(x) + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False