2.9.3 Problem 3

Internal problem ID [18523]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter VII. Linear equations of order higher than the first. section 63. Problems at page 196
Problem number : 3
Date solved : Monday, March 31, 2025 at 05:41:11 PM
CAS classification : [[_3rd_order, _linear, _nonhomogeneous]]

Solved as higher order constant coeff ode

Time used: 0.539 (sec)

The characteristic equation is

$${\lambda }^3-{\lambda }^2+\lambda -1=0$$

The roots of the above equation are

$$\begin{array}{rl}{\lambda }_1& =1\\ {\lambda }_2& =i\\ {\lambda }_3& =-i\end{array}$$

Therefore the homogeneous solution is

$$y_h(x)={\mathrm{e}}^x{c}_1+{\mathrm{e}}^{-ix}{c}_2+{\mathrm{e}}^{ix}{c}_3$$

The fundamental set of solutions for the homogeneous solution are the following

$$\begin{array}{rl}{y}_1& ={\mathrm{e}}^x\\ y_2& ={\mathrm{e}}^{-ix}\\ y_3& ={\mathrm{e}}^{ix}\end{array}$$

This is higher order nonhomogeneous ODE. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE And $y_p$ is a particular solution to the nonhomogeneous ODE. $y_h$ is the solution to

$$y^{\prime \prime \prime }-y^{\prime \prime }+y^{\prime }-y=0$$

Now the particular solution to the given ODE is found

$$y^{\prime \prime \prime }-y^{\prime \prime }+y^{\prime }-y=\cos \left(x\right)$$

Let the particular solution be

$$y_p=U_1{y}_1+U_2{y}_2+U_3{y}_3$$

Where $y_i$ are the basis solutions found above for the homogeneous solution $y_h$ and $U_i(x)$ are functions to be determined as follows

$$U_i=(-1{)}^{n-i}\int \frac{F(x)W_i(x)}{aW(x)}dx$$

Where $W(x)$ is the Wronskian and $W_i(x)$ is the Wronskian that results after deleting the last row and the $i$-th column of the determinant and $n$ is the order of the ODE or equivalently, the number of basis solutions, and $a$ is the coefficient of the leading derivative in the ODE, and $F(x)$ is the RHS of the ODE. Therefore, the first step is to find the Wronskian $W\left(x\right)$. This is given by

$$W(x)=\left|\begin{array}{ccc}{y}_1& y_2& y_3\\ y_1^{\prime }& y_2^{\prime }& y_3^{\prime }\\ y_1^{″}& y_2^{″}& y_3^{″}\end{array}\right|$$

Substituting the fundamental set of solutions $y_i$ found above in the Wronskian gives

$$\begin{array}{rl}W& =\left[\begin{array}{ccc}{\mathrm{e}}^x& {\mathrm{e}}^{-ix}& {\mathrm{e}}^{ix}\\ {\mathrm{e}}^x& -i{\mathrm{e}}^{-ix}& i{\mathrm{e}}^{ix}\\ {\mathrm{e}}^x& -{\mathrm{e}}^{-ix}& -{\mathrm{e}}^{ix}\end{array}\right]\\ |W|& =4i{\mathrm{e}}^x{\mathrm{e}}^{-ix}{\mathrm{e}}^{ix}\end{array}$$

The determinant simplifies to

$$\begin{array}{rl}|W|& =4i{\mathrm{e}}^x\end{array}$$

Now we determine $W_i$ for each $U_i$.

$$\begin{array}{rl}{W}_1(x)& =det\left[\begin{array}{cc}{\mathrm{e}}^{-ix}& {\mathrm{e}}^{ix}\\ -i{\mathrm{e}}^{-ix}& i{\mathrm{e}}^{ix}\end{array}\right]\\ & =2i\end{array}$$

$$\begin{array}{rl}{W}_2(x)& =det\left[\begin{array}{cc}{\mathrm{e}}^x& {\mathrm{e}}^{ix}\\ {\mathrm{e}}^x& i{\mathrm{e}}^{ix}\end{array}\right]\\ & =(-1+i){\mathrm{e}}^{(1+i)x}\end{array}$$

$$\begin{array}{rl}{W}_3(x)& =det\left[\begin{array}{cc}{\mathrm{e}}^x& {\mathrm{e}}^{-ix}\\ {\mathrm{e}}^x& -i{\mathrm{e}}^{-ix}\end{array}\right]\\ & =(-1-i){\mathrm{e}}^{(1-i)x}\end{array}$$

Now we are ready to evaluate each $U_i(x)$.

$$\begin{array}{rl}{U}_1& =(-1{)}^{3-1}\int \frac{F(x)W_1(x)}{aW(x)}dx\\ & =(-1{)}^2\int \frac{(\cos \left(x\right))\left(2i\right)}{\left(1\right)\left(4i{\mathrm{e}}^x\right)}dx\\ & =\int \frac{2i\cos \left(x\right)}{4i{\mathrm{e}}^x}dx\\ & =\int \left(\frac{\cos \left(x\right){\mathrm{e}}^{-x}}{2}\right)dx\\ & =-\frac{\cos \left(x\right){\mathrm{e}}^{-x}}{4}+\frac{{\mathrm{e}}^{-x}\sin \left(x\right)}{4}\end{array}$$

$$\begin{array}{rl}{U}_2& =(-1{)}^{3-2}\int \frac{F(x)W_2(x)}{aW(x)}dx\\ & =(-1{)}^1\int \frac{(\cos \left(x\right))\left((-1+i){\mathrm{e}}^{(1+i)x}\right)}{\left(1\right)\left(4i{\mathrm{e}}^x\right)}dx\\ & =-\int \frac{(-1+i)\cos \left(x\right){\mathrm{e}}^{(1+i)x}}{4i{\mathrm{e}}^x}dx\\ & =-\int ((\frac{1}{4}+\frac{i}{4})\cos \left(x\right){\mathrm{e}}^{ix})dx\\ & =-\frac{x}{8}-\frac{ix}{8}-\frac{{\mathrm{e}}^{2ix}}{16}+\frac{i{\mathrm{e}}^{2ix}}{16}\\ & =-\frac{x}{8}-\frac{ix}{8}-\frac{{\mathrm{e}}^{2ix}}{16}+\frac{i{\mathrm{e}}^{2ix}}{16}\end{array}$$

$$\begin{array}{rl}{U}_3& =(-1{)}^{3-3}\int \frac{F(x)W_3(x)}{aW(x)}dx\\ & =(-1{)}^0\int \frac{(\cos \left(x\right))\left((-1-i){\mathrm{e}}^{(1-i)x}\right)}{\left(1\right)\left(4i{\mathrm{e}}^x\right)}dx\\ & =\int \frac{(-1-i)\cos \left(x\right){\mathrm{e}}^{(1-i)x}}{4i{\mathrm{e}}^x}dx\\ & =\int ((-\frac{1}{4}+\frac{i}{4})\cos \left(x\right){\mathrm{e}}^{-ix})dx\\ & =\int (-\frac{1}{4}+\frac{i}{4})\cos \left(x\right){\mathrm{e}}^{-ix}dx\end{array}$$

Now that all the $U_i$ functions have been determined, the particular solution is found from

$$y_p=U_1{y}_1+U_2{y}_2+U_3{y}_3$$

Hence

$$\begin{array}{rl}{y}_p& =(-\frac{\cos \left(x\right){\mathrm{e}}^{-x}}{4}+\frac{{\mathrm{e}}^{-x}\sin \left(x\right)}{4})\left({\mathrm{e}}^x\right)\\ & +(-\frac{x}{8}-\frac{ix}{8}-\frac{{\mathrm{e}}^{2ix}}{16}+\frac{i{\mathrm{e}}^{2ix}}{16})\left({\mathrm{e}}^{-ix}\right)\\ & +(\int (-\frac{1}{4}+\frac{i}{4})\cos \left(x\right){\mathrm{e}}^{-ix}dx)\left({\mathrm{e}}^{ix}\right)\end{array}$$

Therefore the particular solution is

$$y_p=-\frac{{\mathrm{e}}^{ix}}{8}+(-\frac{3}{16}+\frac{i}{16}){\mathrm{e}}^{-ix}-\frac{x\cos \left(x\right)}{4}-\frac{x\sin \left(x\right)}{4}$$

Which simplifies to

$$y_p=\frac{(-5+i-4x)\cos \left(x\right)}{16}+\frac{(1+i-4x)\sin \left(x\right)}{16}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =({\mathrm{e}}^x{c}_1+{\mathrm{e}}^{-ix}{c}_2+{\mathrm{e}}^{ix}{c}_3)+(\frac{(-5+i-4x)\cos \left(x\right)}{16}+\frac{(1+i-4x)\sin \left(x\right)}{16})\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 33

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

$$y=\frac{(-x+4c_1-2)\cos \left(x\right)}{4}+\frac{(4c_3-x+1)\sin \left(x\right)}{4}+c_2{\mathrm{e}}^x$$

Maple trace

Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
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^2}{d{x}^2}y\left(x\right)-\frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{d}{dx}y\left(x\right)-y\left(x\right)=\cos \left(x\right)\\ \bullet & & \text{Highest derivative means the order of the ODE is}3\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^3-r^2+r-1=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=[1,\mathrm{I},-\mathrm{I}]\\ \bullet & & \text{Homogeneous solution from}r=1\\ & & y_1\left(x\right)={\mathrm{e}}^x\\ \bullet & & \text{Homogeneous solutions from}r=\mathrm{I}\text{and}r=-\mathrm{I}\\ & & [y_2\left(x\right)=\sin \left(x\right),y_3\left(x\right)=\cos \left(x\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)+\mathit{C}\mathit{3}{y}_3\left(x\right)+y_p\left(x\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^x+\mathit{C}\mathit{2}\sin \left(x\right)+\mathit{C}\mathit{3}\cos \left(x\right)+y_p\left(x\right)\\ ◻& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \circ & \text{Define the forcing function of the ODE}\\ & & f\left(x\right)=\cos \left(x\right)\\ & \circ & \text{Form of the particular solution to the ODE where the}{u}_i\left(x\right)\text{are to be found}\\ & & y_p\left(x\right)=\stackrel{3}{\sum _{i=1}}{u}_i\left(x\right)y_i\left(x\right)\\ & \circ & \text{Calculate the 1st derivative of}{y}_p\left(x\right)\\ & & \frac{d}{dx}{y}_p\left(x\right)=\stackrel{3}{\sum _{i=1}}(\left(\frac{d}{dx}{u}_i\left(x\right)\right)y_i\left(x\right)+u_i\left(x\right)\left(\frac{d}{dx}{y}_i\left(x\right)\right))\\ & \circ & \text{Choose equation to add to a system of equations in}\frac{d}{dx}{u}_i\left(x\right)\\ & & \stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)y_i\left(x\right)=0\\ & \circ & \text{Calculate the 2nd derivative of}{y}_p\left(x\right)\\ & & \frac{d}{dx}\frac{d}{dx}{y}_p\left(x\right)=\stackrel{3}{\sum _{i=1}}(\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}{y}_i\left(x\right)\right)+u_i\left(x\right)\left(\frac{d}{dx}\frac{d}{dx}{y}_i\left(x\right)\right))\\ & \circ & \text{Choose equation to add to a system of equations in}\frac{d}{dx}{u}_i\left(x\right)\\ & & \stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}{y}_i\left(x\right)\right)=0\\ & \circ & \text{The ODE is of the following form where the}{P}_i\left(x\right)\text{in this situation are the coefficients of the derivatives in the ODE}\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)+(\stackrel{2}{\sum _{i=0}}{P}_i\left(x\right)\left(\frac{d^i}{d{x}^i}y\left(x\right)\right))=f\left(x\right)\\ & \circ & \text{Substitute}{y}_p\left(x\right)=\stackrel{3}{\sum _{i=1}}{u}_i\left(x\right)y_i\left(x\right)\text{into the ODE}\\ & & (\stackrel{2}{\sum _{j=0}}{P}_j\left(x\right)(\stackrel{3}{\sum _{i=1}}{u}_i\left(x\right)\left(\frac{d^j}{d{x}^j}{y}_i\left(x\right)\right)))+\stackrel{3}{\sum _{i=1}}(\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d}{dx}{y}_i\left(x\right)\right)+u_i\left(x\right)\left(\frac{d}{dx}\frac{d^2}{d{x}^2}{y}_i\left(x\right)\right))=f\left(x\right)\\ & \circ & \text{Rearrange the ODE}\\ & & \stackrel{3}{\sum _{i=1}}(u_i\left(x\right)\cdot ((\stackrel{2}{\sum _{j=0}}{P}_j\left(x\right)\left(\frac{d^j}{d{x}^j}{y}_i\left(x\right)\right))+\frac{d}{dx}\frac{d^2}{d{x}^2}{y}_i\left(x\right))+\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d}{dx}{y}_i\left(x\right)\right))=f\left(x\right)\\ & \circ & \text{Notice that}{y}_i\left(x\right)\text{are solutions to the homogeneous equation so the first term in the sum is 0}\\ & & \stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d}{dx}{y}_i\left(x\right)\right)=f\left(x\right)\\ & \circ & \text{We have now made a system of}3\text{equations in}3\text{unknowns (}\frac{d}{dx}{u}_i\left(x\right))\\ & & [\stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)y_i\left(x\right)=0,\stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}{y}_i\left(x\right)\right)=0,\stackrel{3}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d}{dx}{y}_i\left(x\right)\right)=f\left(x\right)]\\ & \circ & \text{Convert the system to linear algebra format, notice that the matrix is the wronskian}W\\ & & \left[\begin{array}{ccc}{y}_1\left(x\right)& y_2\left(x\right)& y_3\left(x\right)\\ \frac{d}{dx}{y}_1\left(x\right)& \frac{d}{dx}{y}_2\left(x\right)& \frac{d}{dx}{y}_3\left(x\right)\\ \frac{d}{dx}\frac{d}{dx}{y}_1\left(x\right)& \frac{d}{dx}\frac{d}{dx}{y}_2\left(x\right)& \frac{d}{dx}\frac{d}{dx}{y}_3\left(x\right)\end{array}\right]\cdot \left[\begin{array}{c}\frac{d}{dx}{u}_1\left(x\right)\\ \frac{d}{dx}{u}_2\left(x\right)\\ \frac{d}{dx}{u}_3\left(x\right)\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ f\left(x\right)\end{array}\right]\\ & \circ & \text{Solve for the varied parameters}\\ & & \left[\begin{array}{c}{u}_1\left(x\right)\\ u_2\left(x\right)\\ u_3\left(x\right)\end{array}\right]=\int \frac{1}{W}\cdot \left[\begin{array}{c}0\\ 0\\ f\left(x\right)\end{array}\right]dx\\ & \circ & \text{Substitute in the homogeneous solutions and forcing function and solve}\\ & & \left[\begin{array}{c}{u}_1\left(x\right)\\ u_2\left(x\right)\\ u_3\left(x\right)\end{array}\right]=\left[\begin{array}{c}-\frac{{\mathrm{e}}^{-x}\cos \left(x\right)}{4}+\frac{{\mathrm{e}}^{-x}\sin \left(x\right)}{4}\\ -\frac{x}{4}-\frac{\sin \left(2x\right)}{8}+\frac{\cos \left(2x\right)}{8}\\ -\frac{x}{4}-\frac{\sin \left(2x\right)}{8}-\frac{\cos \left(2x\right)}{8}\end{array}\right]\\ & & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & & y_p\left(x\right)=\frac{(-2x-3)\cos \left(x\right)}{8}+\frac{(-2x+1)\sin \left(x\right)}{8}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^x+\mathit{C}\mathit{2}\sin \left(x\right)+\mathit{C}\mathit{3}\cos \left(x\right)+\frac{(-2x-3)\cos \left(x\right)}{8}+\frac{(-2x+1)\sin \left(x\right)}{8}\end{array}$$

✓ Mathematica. Time used: 0.038 (sec). Leaf size: 40

ode=D[y[x],{x,3}]-D[y[x],{x,2}]+D[y[x],x]-y[x]==Cos[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{1}{4}(4c_3{e}^x-(x+2-4c_1)\cos (x)+(-x+1+4c_2)\sin (x))$$

✓ Sympy. Time used: 0.205 (sec). Leaf size: 24

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-y(x) - cos(x) + Derivative(y(x), x) - Derivative(y(x), (x, 2)) + Derivative(y(x), (x, 3)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=C_3{e}^x+(C_1-\frac{x}{4})\sin \left(x\right)+(C_2-\frac{x}{4})\cos \left(x\right)$$