2.3.1 Problem 1

Internal problem ID [8859]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 1
Date solved : Wednesday, March 05, 2025 at 06:56:29 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

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

Solved as second order linear constant coeff ode

Time used: 0.455 (sec)

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

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

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

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

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 $A=1,B=c,C=k$ into the above gives

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

Hence

$$\begin{array}{rl}{\lambda }_1& =-\frac{c}{2}+\frac{\sqrt{c^2-4k}}{2}\\ {\lambda }_2& =-\frac{c}{2}-\frac{\sqrt{c^2-4k}}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =\frac{(1+i+(1-i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2}\\ {\lambda }_2& =\frac{(-1-i+(-1+i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2}\end{array}$$

The roots are complex but they are not conjugate of each others. Hence simplification using Euler relation is not possible here. Therefore the final solution is

$$\begin{array}{rl}y& =c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x}\\ & =c_1{e}^{x(\frac{(1+i+(1-i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2})}+c_2{e}^{x(\frac{(-1-i+(-1+i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2})}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{x(\frac{(1+i+(1-i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2})}+c_2{\mathrm{e}}^{x(\frac{(-1-i+(-1+i)\sqrt{\mathrm{signum}(c^2-4k)})\sqrt{c^2-4k}}{4}-\frac{c}{2})}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.170 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{c^2-4k}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =c^2-4k\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(\frac{c^2}{4}-k)z(x)\end{array}$$

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 2.37: 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 $r=\frac{c^2}{4}-k$ 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

$$z_1(x)={\mathrm{e}}^{\frac{x\sqrt{c^2-4k}}{2}}$$

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

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

Which simplifies to

$$y_1={\mathrm{e}}^{\frac{x(\sqrt{c^2-4k}-c)}{2}}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{c}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-xc}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{{\mathrm{e}}^{-xc}{\mathrm{e}}^{-x(\sqrt{c^2-4k}-c)}}{\sqrt{c^2-4k}})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{\frac{x(\sqrt{c^2-4k}-c)}{2}}\right)+c_2\left({\mathrm{e}}^{\frac{x(\sqrt{c^2-4k}-c)}{2}}(-\frac{{\mathrm{e}}^{-xc}{\mathrm{e}}^{-x(\sqrt{c^2-4k}-c)}}{\sqrt{c^2-4k}})\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{\frac{x(\sqrt{c^2-4k}-c)}{2}}-\frac{c_2{\mathrm{e}}^{-\frac{x(c+\sqrt{c^2-4k})}{2}}}{\sqrt{c^2-4k}}\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 41

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

$$y=c_1{\mathrm{e}}^{\frac{(-c+\sqrt{c^2-4k})x}{2}}+c_2{\mathrm{e}}^{-\frac{(c+\sqrt{c^2-4k})x}{2}}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)+c\left(\frac{d}{dx}y\left(x\right)\right)+ky\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & cr+r^2+k=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{(-c)\pm \left(\sqrt{c^2-4k}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\frac{c}{2}-\frac{\sqrt{c^2-4k}}{2},-\frac{c}{2}+\frac{\sqrt{c^2-4k}}{2})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)={\mathrm{e}}^{(-\frac{c}{2}-\frac{\sqrt{c^2-4k}}{2})x}\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)={\mathrm{e}}^{(-\frac{c}{2}+\frac{\sqrt{c^2-4k}}{2})x}\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{(-\frac{c}{2}-\frac{\sqrt{c^2-4k}}{2})x}+\mathit{C}\mathit{2}{\mathrm{e}}^{(-\frac{c}{2}+\frac{\sqrt{c^2-4k}}{2})x}\end{array}$$

✓ Mathematica. Time used: 7.017 (sec). Leaf size: 2548

ode=D[y[x],{x,3}]-x^3*D[y[x],x]-x^2*y[x]-x^3==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Too large to display

✓ Sympy. Time used: 0.228 (sec). Leaf size: 39

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

$$y\left(x\right)=C_1{e}^{\frac{x(-c+\sqrt{c^2-4k})}{2}}+C_2{e}^{-\frac{x(c+\sqrt{c^2-4k})}{2}}$$