2.2.6 Problem 5 (iiI=i)

Internal problem ID [18441]
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 (iiI=i)
Date solved : Saturday, April 26, 2025 at 11:17:20 AM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.075 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }-4x^{\prime }+5x& =0\end{array}$$

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 $A=1,B=-4,C=5$. Let the solution be $x=e^{\lambda t}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{t\lambda }-4\lambda {\mathrm{e}}^{t\lambda }+5{\mathrm{e}}^{t\lambda }=0\end{array}$$

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

$$\begin{array}{}\text{(2)}& {\lambda }^2-4\lambda +5=0\end{array}$$

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 $A=1,B=-4,C=5$ into the above gives

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

Hence

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

Which simplifies to

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

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

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

Where $\alpha =2$ and $\beta =1$. Therefore the final solution, when using Euler relation, can be written as

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

Which becomes

$$x=e^{2t}(c_1\cos \left(t\right)+c_2\sin \left(t\right))$$

Will add steps showing solving for IC soon.

Summary of solutions found

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

Figure 2.54: Slope field $x^{\prime \prime }-4x^{\prime }+5x=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.191 (sec)

Solve

$$\begin{array}{rl}{x}^{\prime \prime }-4x^{\prime }+5x& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& x^{\prime \prime }-4x^{\prime }+5x& =0\text{(2)}& A{x}^{\prime \prime }+B{x}^{\prime }+Cx& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =-4C& =5\end{array}$$

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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 2.12: 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 $r=-1$ 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

$$z_1(t)=\cos \left(t\right)$$

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

$$\begin{array}{rl}{x}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-4}{1}dt}\\ & =z_1{e}^{2t}\\ & =z_1\left({\mathrm{e}}^{2t}\right)\end{array}$$

Which simplifies to

$$x_1={\mathrm{e}}^{2t}\cos \left(t\right)$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Figure 2.55: Slope field $x^{\prime \prime }-4x^{\prime }+5x=0$

✓ Maple. Time used: 0.001 (sec). Leaf size: 18

ode:=diff(diff(x(t),t),t)-4*diff(x(t),t)+5*x(t) = 0; 
dsolve(ode,x(t), singsol=all);
 

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

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dt}\frac{d}{dt}x\left(t\right)-4\frac{d}{dt}x\left(t\right)+5x\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}x\left(t\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2-4r+5=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{4\pm \left(\sqrt{-4}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(2-\mathrm{I},2+\mathrm{I})\\ \bullet & & \text{1st solution of the ODE}\\ & & x_1\left(t\right)={\mathrm{e}}^{2t}\cos \left(t\right)\\ \bullet & & \text{2nd solution of the ODE}\\ & & x_2\left(t\right)={\mathrm{e}}^{2t}\sin \left(t\right)\\ \bullet & & \text{General solution of the ODE}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{x}_1\left(t\right)+\mathit{C}\mathit{2}{x}_2\left(t\right)\\ \bullet & & \text{Substitute in solutions}\\ & & x\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{2t}\cos \left(t\right)+\mathit{C}\mathit{2}{\mathrm{e}}^{2t}\sin \left(t\right)\end{array}$$

✓ Mathematica. Time used: 0.015 (sec). Leaf size: 22

ode=D[x[t],{t,2}]-4*D[x[t],t]+5*x[t]==0; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

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

✓ Sympy. Time used: 0.146 (sec). Leaf size: 17

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(5*x(t) - 4*Derivative(x(t), t) + Derivative(x(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

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