2.8.2 Problem 2

Internal problem ID [19763]
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 : 2
Date solved : Friday, November 28, 2025 at 06:51:35 PM
CAS classification : [[_high_order, _missing_y]]

Solve

\begin{align*} y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }&={\mathrm e}^{2 x} \\ \end{align*}

Solved as higher order constant coeff ode

Time used: 0.165 (sec)

The characteristic equation is

\[ \lambda ^{4}-3 \lambda ^{3}+3 \lambda ^{2}-\lambda = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= 0\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end{align*}

Therefore the homogeneous solution is

\[ y_h(x)=c_1 +c_2 \,{\mathrm e}^{x}+x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4 \]

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

\begin{align*} y_1 &= 1\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= x \,{\mathrm e}^{x}\\ y_4 &= x^{2} {\mathrm e}^{x} \end{align*}

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

\[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime } = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime } = {\mathrm e}^{2 x} \]

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

Looking at the RHS of the ode, which is

\[ {\mathrm e}^{2 x} \]

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

\[ [\{{\mathrm e}^{2 x}\}] \]

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

\[ \{1, x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\} \]

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

\[ y_p = A_{1} {\mathrm e}^{2 x} \]

The unknowns \(\{A_{1}\}\) 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

\[ 2 A_{1} {\mathrm e}^{2 x} = {\mathrm e}^{2 x} \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{2}}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {{\mathrm e}^{2 x}}{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 +c_2 \,{\mathrm e}^{x}+x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4\right ) + \left (\frac {{\mathrm e}^{2 x}}{2}\right ) \\ \end{align*}

✓ Maple. Time used: 0.001 (sec). Leaf size: 33

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

\[ y = \frac {{\mathrm e}^{2 x}}{2}+\left (\left (x^{2}-2 x +2\right ) c_3 +c_2 x +c_1 -c_2 \right ) {\mathrm e}^{x}+c_4 \]

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] 
-> Calling odsolve with the ODE, diff(diff(diff(_b(_a),_a),_a),_a) = 3*diff( 
diff(_b(_a),_a),_a)-3*diff(_b(_a),_a)+_b(_a)+exp(2*_a), _b(_a) 
   *** Sublevel 2 *** 
   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 
<- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )-3 \frac {d^{3}}{d x^{3}}y \left (x \right )+3 \frac {d^{2}}{d x^{2}}y \left (x \right )-\frac {d}{d x}y \left (x \right )={\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{4}-3 r^{3}+3 r^{2}-r =0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial and corresponding multiplicities}\hspace {3pt} \\ {} & {} & r =\left [\left [0, 1\right ], \left [1, 3\right ]\right ] \\ \bullet & {} & \textrm {Homogeneous solution from}\hspace {3pt} r =0 \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {1st homogeneous solution from}\hspace {3pt} r =1 \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {2nd homogeneous solution from}\hspace {3pt} r =1 \\ {} & {} & y_{3}\left (x \right )=x \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {3rd homogeneous solution from}\hspace {3pt} r =1 \\ {} & {} & y_{4}\left (x \right )=x^{2} {\mathrm e}^{x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right )+\mathit {C3} y_{3}\left (x \right )+\mathit {C4} y_{4}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} +\mathit {C2} \,{\mathrm e}^{x}+\mathit {C3} x \,{\mathrm e}^{x}+\mathit {C4} \,x^{2} {\mathrm e}^{x}+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Define the forcing function of the ODE}\hspace {3pt} \\ {} & {} & f \left (x \right )={\mathrm e}^{2 x} \\ {} & \circ & \textrm {Form of the particular solution to the ODE where the}\hspace {3pt} u_{i}\left (x \right )\hspace {3pt}\textrm {are to be found}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (x \right ) y_{i}\left (x \right ) \\ {} & \circ & \textrm {Calculate the 1st derivative of}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & \frac {d}{d x}y_{p}\left (x \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d x}u_{i}\left (x \right )\right ) y_{i}\left (x \right )+u_{i}\left (x \right ) \left (\frac {d}{d x}y_{i}\left (x \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d x}u_{i}\left (x \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) y_{i}\left (x \right )=0 \\ {} & \circ & \textrm {Calculate the 2nd derivative of}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & \frac {d^{2}}{d x^{2}}y_{p}\left (x \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d}{d x}y_{i}\left (x \right )\right )+u_{i}\left (x \right ) \left (\frac {d^{2}}{d x^{2}}y_{i}\left (x \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d x}u_{i}\left (x \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d}{d x}y_{i}\left (x \right )\right )=0 \\ {} & \circ & \textrm {Calculate the 3rd derivative of}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & \frac {d^{3}}{d x^{3}}y_{p}\left (x \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{2}}{d x^{2}}y_{i}\left (x \right )\right )+u_{i}\left (x \right ) \left (\frac {d^{3}}{d x^{3}}y_{i}\left (x \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d x}u_{i}\left (x \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{2}}{d x^{2}}y_{i}\left (x \right )\right )=0 \\ {} & \circ & \textrm {The ODE is of the following form where the}\hspace {3pt} P_{i}\left (x \right )\hspace {3pt}\textrm {in this situation are the coefficients of the derivatives in the ODE}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )+\left (\moverset {3}{\munderset {i =0}{\sum }}P_{i}\left (x \right ) \left (\frac {d^{i}}{d x^{i}}y \left (x \right )\right )\right )=f \left (x \right ) \\ {} & \circ & \textrm {Substitute}\hspace {3pt} y_{p}\left (x \right )=\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (x \right ) y_{i}\left (x \right )\hspace {3pt}\textrm {into the ODE}\hspace {3pt} \\ {} & {} & \left (\moverset {3}{\munderset {j =0}{\sum }}P_{j}\left (x \right ) \left (\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (x \right ) \left (\frac {d^{j}}{d x^{j}}y_{i}\left (x \right )\right )\right )\right )+\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{3}}{d x^{3}}y_{i}\left (x \right )\right )+u_{i}\left (x \right ) \left (\frac {d^{4}}{d x^{4}}y_{i}\left (x \right )\right )\right )=f \left (x \right ) \\ {} & \circ & \textrm {Rearrange the ODE}\hspace {3pt} \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (u_{i}\left (x \right )\cdot \left (\left (\moverset {3}{\munderset {j =0}{\sum }}P_{j}\left (x \right ) \left (\frac {d^{j}}{d x^{j}}y_{i}\left (x \right )\right )\right )+\frac {d^{4}}{d x^{4}}y_{i}\left (x \right )\right )+\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{3}}{d x^{3}}y_{i}\left (x \right )\right )\right )=f \left (x \right ) \\ {} & \circ & \textrm {Notice that}\hspace {3pt} y_{i}\left (x \right )\hspace {3pt}\textrm {are solutions to the homogeneous equation so the first term in the sum is 0}\hspace {3pt} \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{3}}{d x^{3}}y_{i}\left (x \right )\right )=f \left (x \right ) \\ {} & \circ & \textrm {We have now made a system of}\hspace {3pt} 4\hspace {3pt}\textrm {equations in}\hspace {3pt} 4\hspace {3pt}\textrm {unknowns (}\hspace {3pt} \frac {d}{d x}u_{i}\left (x \right )) \\ {} & {} & \left [\moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) y_{i}\left (x \right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d}{d x}y_{i}\left (x \right )\right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{2}}{d x^{2}}y_{i}\left (x \right )\right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d x}u_{i}\left (x \right )\right ) \left (\frac {d^{3}}{d x^{3}}y_{i}\left (x \right )\right )=f \left (x \right )\right ] \\ {} & \circ & \textrm {Convert the system to linear algebra format, notice that the matrix is the wronskian}\hspace {3pt} 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}{d x}y_{1}\left (x \right ) & \frac {d}{d x}y_{2}\left (x \right ) & \frac {d}{d x}y_{3}\left (x \right ) & \frac {d}{d x}y_{4}\left (x \right ) \\ \frac {d^{2}}{d x^{2}}y_{1}\left (x \right ) & \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) & \frac {d^{2}}{d x^{2}}y_{3}\left (x \right ) & \frac {d^{2}}{d x^{2}}y_{4}\left (x \right ) \\ \frac {d^{3}}{d x^{3}}y_{1}\left (x \right ) & \frac {d^{3}}{d x^{3}}y_{2}\left (x \right ) & \frac {d^{3}}{d x^{3}}y_{3}\left (x \right ) & \frac {d^{3}}{d x^{3}}y_{4}\left (x \right ) \end {array}\right ]\cdot \left [\begin {array}{c} \frac {d}{d x}u_{1}\left (x \right ) \\ \frac {d}{d x}u_{2}\left (x \right ) \\ \frac {d}{d x}u_{3}\left (x \right ) \\ \frac {d}{d x}u_{4}\left (x \right ) \end {array}\right ]=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \left (x \right ) \end {array}\right ] \\ {} & \circ & \textrm {Solve for the varied parameters}\hspace {3pt} \\ {} & {} & \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 W^{-1}\cdot \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \left (x \right ) \end {array}\right ]d x \\ {} & \circ & \textrm {Substitute in the homogeneous solutions and forcing function and solve}\hspace {3pt} \\ {} & {} & \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 {{\mathrm e}^{2 x}}{2} \\ \frac {\left (x^{2}+2\right ) {\mathrm e}^{2 x}}{2 \,{\mathrm e}^{x}} \\ -\frac {x \,{\mathrm e}^{2 x}}{{\mathrm e}^{x}} \\ \frac {{\mathrm e}^{2 x}}{2 \,{\mathrm e}^{x}} \end {array}\right ] \\ & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {{\mathrm e}^{2 x}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} +\mathit {C2} \,{\mathrm e}^{x}+\mathit {C3} x \,{\mathrm e}^{x}+\mathit {C4} \,x^{2} {\mathrm e}^{x}+\frac {{\mathrm e}^{2 x}}{2} \end {array} \]

✓ Mathematica. Time used: 0.032 (sec). Leaf size: 41

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

\begin{align*} y(x)&\to \frac {1}{2} e^x \left (e^x+2 \left (c_3 \left (x^2-2 x+2\right )+c_2 (x-1)+c_1\right )\right )+c_4 \end{align*}

✓ Sympy. Time used: 0.122 (sec). Leaf size: 22

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

\[ y{\left (x \right )} = C_{1} + \left (C_{2} + x \left (C_{3} + C_{4} x\right )\right ) e^{x} + \frac {e^{2 x}}{2} \]