2.2.4 Problem 5 (i)

Internal problem ID [18446]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 4. Autonomous systems. Exercises at page 69
Problem number : 5 (i)
Date solved : Monday, March 31, 2025 at 05:29:32 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.060 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }-5x^{\prime }+6x& =0\end{array}$$

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

$$A{x}^{″}(t)+B{x}^{\prime }(t)+Cx(t)=0$$

Where in the above $A=1,B=-5,C=6$. Let the solution be $x=e^{\lambda t}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{t\lambda }-5\lambda {\mathrm{e}}^{t\lambda }+6{\mathrm{e}}^{t\lambda }=0\end{array}$$

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

$$\begin{array}{}\text{(2)}& {\lambda }^2-5\lambda +6=0\end{array}$$

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 $A=1,B=-5,C=6$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{5}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{-5^2-(4)\left(1\right)\left(6\right)}\\ & =\frac{5}{2}\pm \frac{1}{2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =\frac{5}{2}+\frac{1}{2}\\ {\lambda }_2& =\frac{5}{2}-\frac{1}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =3\\ {\lambda }_2& =2\end{array}$$

Since roots are real and distinct, then the solution is

$$\begin{array}{rl}x& =c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t}\\ x& =c_1{e}^{\left(3\right)t}+c_2{e}^{\left(2\right)t}\end{array}$$

Or

$$x=c_1{\mathrm{e}}^{3t}+c_2{\mathrm{e}}^{2t}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =c_1{\mathrm{e}}^{3t}+c_2{\mathrm{e}}^{2t}\end{array}$$

Figure 2.50: Slope field $x^{\prime \prime }-5x^{\prime }+6x=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.071 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }-5x^{\prime }+6x& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^{\prime \prime }-5x^{\prime }+6x& =0\text{(2)}& A{x}^{\prime \prime }+B{x}^{\prime }+Cx& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =-5C& =6\end{array}$$

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{1}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =4\end{array}$$

Therefore eq. (4) becomes

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

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

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

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 2.10: 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 $r=\frac{1}{4}$ is not a function of $t$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(t)={\mathrm{e}}^{-\frac{t}{2}}$$

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

$$\begin{array}{rl}{x}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-5}{1}dt}\\ & =z_1{e}^{\frac{5t}{2}}\\ & =z_1\left({\mathrm{e}}^{\frac{5t}{2}}\right)\end{array}$$

Which simplifies to

$$x_1={\mathrm{e}}^{2t}$$

The second solution $x_2$ to the original ode is found using reduction of order

$$x_2=x_1\int \frac{e^{\int -\frac{B}{A}dt}}{x_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{x}_2& =x_1\int \frac{e^{\int -\frac{-5}{1}dt}}{{\left(x_1\right)}^2}dt\\ & =x_1\int \frac{e^{5t}}{{\left(x_1\right)}^2}dt\\ & =x_1\left({\mathrm{e}}^{5t}{\mathrm{e}}^{-4t}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}x& =c_1{x}_1+c_2{x}_2\\ & =c_1\left({\mathrm{e}}^{2t}\right)+c_2\left({\mathrm{e}}^{2t}\left({\mathrm{e}}^{5t}{\mathrm{e}}^{-4t}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}x& =c_1{\mathrm{e}}^{2t}+c_2{\mathrm{e}}^{3t}\end{array}$$

Figure 2.51: Slope field $x^{\prime \prime }-5x^{\prime }+6x=0$

✓ Maple. Time used: 0.003 (sec). Leaf size: 15

ode:=diff(diff(x(t),t),t)-5*diff(x(t),t)+6*x(t) = 0; 
dsolve(ode,x(t), singsol=all);
 

$$x=(c_1{\mathrm{e}}^t+c_2){\mathrm{e}}^{2t}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dt}\frac{d}{dt}x\left(t\right)-5\frac{d}{dt}x\left(t\right)+6x\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}x\left(t\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2-5r+6=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & (r-2)(r-3)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(2,3)\\ \bullet & & \text{1st solution of the ODE}\\ & & x_1\left(t\right)={\mathrm{e}}^{2t}\\ \bullet & & \text{2nd solution of the ODE}\\ & & x_2\left(t\right)={\mathrm{e}}^{3t}\\ \bullet & & \text{General solution of the ODE}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{x}_1\left(t\right)+\mathit{C}\mathit{2}{x}_2\left(t\right)\\ \bullet & & \text{Substitute in solutions}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{2t}+\mathit{C}\mathit{2}{\mathrm{e}}^{3t}\end{array}$$

✓ Mathematica. Time used: 0.013 (sec). Leaf size: 20

ode=D[x[t],{t,2}]-5*D[x[t],t]+6*x[t]==0; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

$$x(t)\to e^{2t}(c_2{e}^t+c_1)$$

✓ Sympy. Time used: 0.158 (sec). Leaf size: 14

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(6*x(t) - 5*Derivative(x(t), t) + Derivative(x(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

$$x\left(t\right)=(C_1+C_2{e}^t)e^{2t}$$