2.2.42 Problem 42

Internal problem ID [9164]
Book : Second order enumerated odes
Section : section 2
Problem number : 42
Date solved : Friday, April 25, 2025 at 05:58:43 PM
CAS classification : [[_high_order, _linear, _nonhomogeneous]]

Solved as higher order constant coeff ode

Time used: 0.163 (sec)

The characteristic equation is

$${\lambda }^4-{\lambda }^3-3{\lambda }^2+5\lambda -2=0$$

The roots of the above equation are

$$\begin{array}{rl}{\lambda }_1& =-2\\ {\lambda }_2& =1\\ {\lambda }_3& =1\\ {\lambda }_4& =1\end{array}$$

Therefore the homogeneous solution is

$$y_h(x)={\mathrm{e}}^x{c}_1+x{\mathrm{e}}^x{c}_2+x^2{\mathrm{e}}^x{c}_3+{\mathrm{e}}^{-2x}{c}_4$$

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

$$\begin{array}{rl}{y}_1& ={\mathrm{e}}^x\\ y_2& =x{\mathrm{e}}^x\\ y_3& =x^2{\mathrm{e}}^x\\ y_4& ={\mathrm{e}}^{-2x}\end{array}$$

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

$$y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3y^{\prime \prime }+5y^{\prime }-2y=0$$

Now the particular solution to the given ODE is found

$$y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3y^{\prime \prime }+5y^{\prime }-2y=x{\mathrm{e}}^x+3{\mathrm{e}}^{-2x}$$

The particular solution is now found using the method of undetermined coefficients.

Looking at the RHS of the ode, which is

$$x{\mathrm{e}}^x+3{\mathrm{e}}^{-2x}$$

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

$$[\{{\mathrm{e}}^{-2x}\},\{x{\mathrm{e}}^x,{\mathrm{e}}^x\}]$$

While the set of the basis functions for the homogeneous solution found earlier is

$$\{x{\mathrm{e}}^x,x^2{\mathrm{e}}^x,{\mathrm{e}}^x,{\mathrm{e}}^{-2x}\}$$

Since ${\mathrm{e}}^x$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{{\mathrm{e}}^{-2x}\},\{x{\mathrm{e}}^x,x^2{\mathrm{e}}^x\}]$$

Since $x{\mathrm{e}}^x$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{{\mathrm{e}}^{-2x}\},\{x^2{\mathrm{e}}^x,x^3{\mathrm{e}}^x\}]$$

Since $x^2{\mathrm{e}}^x$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{{\mathrm{e}}^{-2x}\},\{x^3{\mathrm{e}}^x,x^4{\mathrm{e}}^x\}]$$

Since ${\mathrm{e}}^{-2x}$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{x{\mathrm{e}}^{-2x}\},\{x^3{\mathrm{e}}^x,x^4{\mathrm{e}}^x\}]$$

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

$$y_p=A_{1}x{\mathrm{e}}^{-2x}+A_2{x}^3{\mathrm{e}}^x+A_3{x}^4{\mathrm{e}}^x$$

The unknowns $\{A_1,A_2,A_3\}$ are found by substituting the above trial solution $y_p$ into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

$$24A_3{\mathrm{e}}^x-27A_1{\mathrm{e}}^{-2x}+18A_2{\mathrm{e}}^x+72A_{3}x{\mathrm{e}}^x=x{\mathrm{e}}^x+3{\mathrm{e}}^{-2x}$$

Solving for the unknowns by comparing coefficients results in

$$[A_1=-\frac{1}{9},A_2=-\frac{1}{54},A_3=\frac{1}{72}]$$

Substituting the above back in the above trial solution $y_p$, gives the particular solution

$$y_p=-\frac{x{\mathrm{e}}^{-2x}}{9}-\frac{x^3{\mathrm{e}}^x}{54}+\frac{x^4{\mathrm{e}}^x}{72}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =({\mathrm{e}}^x{c}_1+x{\mathrm{e}}^x{c}_2+x^2{\mathrm{e}}^x{c}_3+{\mathrm{e}}^{-2x}{c}_4)+(-\frac{x{\mathrm{e}}^{-2x}}{9}-\frac{x^3{\mathrm{e}}^x}{54}+\frac{x^4{\mathrm{e}}^x}{72})\end{array}$$

✓ Maple. Time used: 0.006 (sec). Leaf size: 52

ode:=diff(diff(diff(diff(y(x),x),x),x),x)-diff(diff(diff(y(x),x),x),x)-3*diff(diff(y(x),x),x)+5*diff(y(x),x)-2*y(x) = x*exp(x)+3*exp(-2*x); 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{(-216x+1944c_2-216){\mathrm{e}}^{-2x}}{1944}+\frac{{\mathrm{e}}^x(x^4-\frac{4x^3}{3}+(72c_4+\frac{4}{3})x^2+(72c_3-\frac{8}{9})x+72c_1+\frac{8}{27})}{72}$$

Maple trace

Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 4; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 4; 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}{dx}\frac{d^2}{d{x}^2}y\left(x\right)-\frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)-3\frac{d}{dx}\frac{d}{dx}y\left(x\right)+5\frac{d}{dx}y\left(x\right)-2y\left(x\right)=x{\mathrm{e}}^x+3{\mathrm{e}}^{-2x}\\ \bullet & & \text{Highest derivative means the order of the ODE is}4\\ & & \frac{d}{dx}\frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^4-r^3-3r^2+5r-2=0\\ \bullet & & \text{Roots of the characteristic polynomial and corresponding multiplicities}\\ & & r=[[-2,1],[1,3]]\\ \bullet & & \text{Homogeneous solution from}r=-2\\ & & y_1\left(x\right)={\mathrm{e}}^{-2x}\\ \bullet & & \text{1st homogeneous solution from}r=1\\ & & y_2\left(x\right)={\mathrm{e}}^x\\ \bullet & & \text{2nd homogeneous solution from}r=1\\ & & y_3\left(x\right)=x{\mathrm{e}}^x\\ \bullet & & \text{3rd homogeneous solution from}r=1\\ & & y_4\left(x\right)=x^2{\mathrm{e}}^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)+\mathit{C}\mathit{3}{y}_3\left(x\right)+\mathit{C}\mathit{4}{y}_4\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}}^{-2x}+\mathit{C}\mathit{2}{\mathrm{e}}^x+\mathit{C}\mathit{3}x{\mathrm{e}}^x+\mathit{C}\mathit{4}{x}^2{\mathrm{e}}^x+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)=x{\mathrm{e}}^x+3{\mathrm{e}}^{-2x}\\ & \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{4}{\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{4}{\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{4}{\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{4}{\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{4}{\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{Calculate the 3rd derivative of}{y}_p\left(x\right)\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}{y}_p\left(x\right)=\stackrel{4}{\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))\\ & \circ & \text{Choose equation to add to a system of equations in}\frac{d}{dx}{u}_i\left(x\right)\\ & & \stackrel{4}{\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)=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}{dx}\frac{d^2}{d{x}^2}y\left(x\right)+(\stackrel{3}{\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{4}{\sum _{i=1}}{u}_i\left(x\right)y_i\left(x\right)\text{into the ODE}\\ & & (\stackrel{3}{\sum _{j=0}}{P}_j\left(x\right)(\stackrel{4}{\sum _{i=1}}{u}_i\left(x\right)\left(\frac{d^j}{d{x}^j}{y}_i\left(x\right)\right)))+\stackrel{4}{\sum _{i=1}}(\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d^2}{d{x}^2}{y}_i\left(x\right)\right)+u_i\left(x\right)\left(\frac{d}{dx}\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{4}{\sum _{i=1}}(u_i\left(x\right)\cdot ((\stackrel{3}{\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}{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^2}{d{x}^2}{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{4}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d^2}{d{x}^2}{y}_i\left(x\right)\right)=f\left(x\right)\\ & \circ & \text{We have now made a system of}4\text{equations in}4\text{unknowns (}\frac{d}{dx}{u}_i\left(x\right))\\ & & [\stackrel{4}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)y_i\left(x\right)=0,\stackrel{4}{\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{4}{\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)=0,\stackrel{4}{\sum _{i=1}}\left(\frac{d}{dx}{u}_i\left(x\right)\right)\left(\frac{d}{dx}\frac{d^2}{d{x}^2}{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}{cccc}{y}_1\left(x\right)& y_2\left(x\right)& y_3\left(x\right)& y_4\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}{y}_4\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)& \frac{d}{dx}\frac{d}{dx}{y}_4\left(x\right)\\ \frac{d}{dx}\frac{d^2}{d{x}^2}{y}_1\left(x\right)& \frac{d}{dx}\frac{d^2}{d{x}^2}{y}_2\left(x\right)& \frac{d}{dx}\frac{d^2}{d{x}^2}{y}_3\left(x\right)& \frac{d}{dx}\frac{d^2}{d{x}^2}{y}_4\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)\\ \frac{d}{dx}{u}_4\left(x\right)\end{array}\right]=\left[\begin{array}{c}0\\ 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)\\ u_4\left(x\right)\end{array}\right]=\int \frac{1}{W}\cdot \left[\begin{array}{c}0\\ 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)\\ u_4\left(x\right)\end{array}\right]=\left[\begin{array}{c}-\frac{x}{9}-\frac{{\left({\mathrm{e}}^x\right)}^{3}x}{81}+\frac{{\left({\mathrm{e}}^x\right)}^3}{243}\\ \frac{x^3}{27}+\frac{x^2}{54}+\frac{x^4}{24}-\frac{1}{9{\left({\mathrm{e}}^x\right)}^3}-\frac{2x}{9{\left({\mathrm{e}}^x\right)}^3}-\frac{x^2}{6{\left({\mathrm{e}}^x\right)}^3}\\ -\frac{x^2}{18}-\frac{x^3}{9}+\frac{2}{9{\left({\mathrm{e}}^x\right)}^3}+\frac{x}{3{\left({\mathrm{e}}^x\right)}^3}\\ \frac{x^2}{12}-\frac{1}{6{\left({\mathrm{e}}^x\right)}^3}\end{array}\right]\\ & & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & & y_p\left(x\right)=-\frac{{\mathrm{e}}^{-2x}(1+x)}{9}+\frac{(x^4-\frac{4}{3}{x}^3+\frac{4}{3}{x}^2-\frac{8}{9}x+\frac{8}{27}){\mathrm{e}}^x}{72}\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-2x}+\mathit{C}\mathit{2}{\mathrm{e}}^x+\mathit{C}\mathit{3}x{\mathrm{e}}^x+\mathit{C}\mathit{4}{x}^2{\mathrm{e}}^x-\frac{{\mathrm{e}}^{-2x}(1+x)}{9}+\frac{(x^4-\frac{4}{3}{x}^3+\frac{4}{3}{x}^2-\frac{8}{9}x+\frac{8}{27}){\mathrm{e}}^x}{72}\end{array}$$

✓ Mathematica. Time used: 0.279 (sec). Leaf size: 170

ode=D[y[x],{x,4}]-D[y[x],{x,3}]-3*D[y[x],{x,2}]+5*D[y[x],x]-2*y[x]==x*Exp[x]+3*Exp[-2*x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to e^{x}x{\int }_1^x-\frac{1}{9}{e}^{-3K[3]}(3K[3]+1)(e^{3K[3]}K[3]+3)dK[3]+e^{-2x}{\int }_1^x(-\frac{1}{27}{e}^{3K[1]}K[1]-\frac{1}{9})dK[1]+e^x{\int }_1^x\frac{1}{54}{e}^{-3K[2]}(e^{3K[2]}K[2]+3)(9K[2{]}^2+6K[2]+2)dK[2]+\frac{e^x{x}^4}{12}-\frac{1}{6}{e}^{-2x}{x}^2+c_4{e}^x{x}^2+c_3{e}^{x}x+c_1{e}^{-2x}+c_2{e}^x$$

✓ Sympy. Time used: 0.475 (sec). Leaf size: 32

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x*exp(x) - 2*y(x) + 5*Derivative(y(x), x) - 3*Derivative(y(x), (x, 2)) - Derivative(y(x), (x, 3)) + Derivative(y(x), (x, 4)) - 3*exp(-2*x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=(C_1-\frac{x}{9})e^{-2x}+(C_2+x(C_3+x(C_4+\frac{x^2}{72}-\frac{x}{54})))e^x$$