2.1.21 Problem 21

Internal problem ID [9091]
Book : Second order enumerated odes
Section : section 1
Problem number : 21
Date solved : Friday, April 25, 2025 at 05:41:29 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.105 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }+y& =0\end{array}$$

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$$

Where in the above $A=1,B=1,C=1$. Let the solution be $y=e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }+\lambda {\mathrm{e}}^{x\lambda }+{\mathrm{e}}^{x\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda x}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+\lambda +1=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=1,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{-1}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{1^2-(4)\left(1\right)\left(1\right)}\\ & =-\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =-\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =-\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =-\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =-\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =-\frac{1}{2}$ and $\beta =\frac{\sqrt{3}}{2}$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$$

Which becomes

$$y=e^{-\frac{x}{2}}(c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right))$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& ={\mathrm{e}}^{-\frac{x}{2}}(c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right))\end{array}$$

Figure 2.47: Slope field $y^{\prime \prime }+y^{\prime }+y=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.288 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }+y& =0\end{array}$$

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =1C& =1\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-3}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-3\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =-\frac{3z\left(x\right)}{4}\end{array}$$

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 2.8: Necessary conditions for each Kovacic case

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

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

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 $0$ then the necessary conditions for case one are met. Therefore

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

Since $r=-\frac{3}{4}$ 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

$$z_1(x)=\cos \left(\frac{\sqrt{3}x}{2}\right)$$

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

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

Which simplifies to

$$y_1={\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{1}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-x}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{2\sqrt{3}\tan \left(\frac{\sqrt{3}x}{2}\right)}{3}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1({\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right))+c_2({\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)\left(\frac{2\sqrt{3}\tan \left(\frac{\sqrt{3}x}{2}\right)}{3}\right))\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)+\frac{2c_2{\mathrm{e}}^{-\frac{x}{2}}\sqrt{3}\sin \left(\frac{\sqrt{3}x}{2}\right)}{3}\end{array}$$

Figure 2.48: Slope field $y^{\prime \prime }+y^{\prime }+y=0$

✓ Maple. Time used: 0.002 (sec). Leaf size: 28

ode:=diff(diff(y(x),x),x)+diff(y(x),x)+y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y={\mathrm{e}}^{-\frac{x}{2}}(c_1\sin \left(\frac{\sqrt{3}x}{2}\right)+c_2\cos \left(\frac{\sqrt{3}x}{2}\right))$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{d}{dx}y\left(x\right)+y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2+r+1=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{(-1)\pm \left(\sqrt{-3}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\frac{1}{2}-\frac{\mathrm{I}\sqrt{3}}{2},-\frac{1}{2}+\frac{\mathrm{I}\sqrt{3}}{2})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)={\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)={\mathrm{e}}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)+\mathit{C}\mathit{2}{\mathrm{e}}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)\end{array}$$

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 42

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

$$y(x)\to e^{-x/2}(c_2\cos \left(\frac{\sqrt{3}x}{2}\right)+c_1\sin \left(\frac{\sqrt{3}x}{2}\right))$$

✓ Sympy. Time used: 0.146 (sec). Leaf size: 31

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

$$y\left(x\right)=(C_1\sin \left(\frac{\sqrt{3}x}{2}\right)+C_2\cos \left(\frac{\sqrt{3}x}{2}\right))e^{-\frac{x}{2}}$$