2.7.10 Problem 11

Internal problem ID [19759]
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 56. Problems at page 163
Problem number : 11
Date solved : Friday, November 28, 2025 at 06:51:03 PM
CAS classification : [[_high_order, _missing_y]]

Solve

\begin{align*} x^{\prime \prime \prime \prime }-6 x^{\prime \prime \prime }+11 x^{\prime \prime }-6 x^{\prime }&={\mathrm e}^{-3 t} \\ \end{align*}

Solved as higher order constant coeff ode

Time used: 0.167 (sec)

The characteristic equation is

\[ \lambda ^{4}-6 \lambda ^{3}+11 \lambda ^{2}-6 \lambda = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= 0\\ \lambda _2 &= 1\\ \lambda _3 &= 2\\ \lambda _4 &= 3 \end{align*}

Therefore the homogeneous solution is

\[ x_h(t)=c_2 \,{\mathrm e}^{t}+c_1 +c_3 \,{\mathrm e}^{2 t}+c_4 \,{\mathrm e}^{3 t} \]

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

\begin{align*} x_1 &= 1\\ x_2 &= {\mathrm e}^{t}\\ x_3 &= {\mathrm e}^{2 t}\\ x_4 &= {\mathrm e}^{3 t} \end{align*}

This is higher order nonhomogeneous ODE. Let the solution be

\[ x = x_h + x_p \]

Where \(x_h\) is the solution to the homogeneous ODE And \(x_p\) is a particular solution to the nonhomogeneous ODE. \(x_h\) is the solution to

\[ x^{\prime \prime \prime \prime }-6 x^{\prime \prime \prime }+11 x^{\prime \prime }-6 x^{\prime } = 0 \]

Now the particular solution to the given ODE is found

\[ x^{\prime \prime \prime \prime }-6 x^{\prime \prime \prime }+11 x^{\prime \prime }-6 x^{\prime } = {\mathrm e}^{-3 t} \]

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

Looking at the RHS of the ode, which is

\[ {\mathrm e}^{-3 t} \]

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

\[ [\{{\mathrm e}^{-3 t}\}] \]

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

\[ \{1, {\mathrm e}^{t}, {\mathrm e}^{2 t}, {\mathrm e}^{3 t}\} \]

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.

\[ x_p = A_{1} {\mathrm e}^{-3 t} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(x_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 360 A_{1} {\mathrm e}^{-3 t} = {\mathrm e}^{-3 t} \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{360}}\right ] \]

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

\[ x_p = \frac {{\mathrm e}^{-3 t}}{360} \]

Therefore the general solution is

\begin{align*} x &= x_h + x_p \\ &= \left (c_2 \,{\mathrm e}^{t}+c_1 +c_3 \,{\mathrm e}^{2 t}+c_4 \,{\mathrm e}^{3 t}\right ) + \left (\frac {{\mathrm e}^{-3 t}}{360}\right ) \\ \end{align*}

✓ Maple. Time used: 0.002 (sec). Leaf size: 30

ode:=diff(diff(diff(diff(x(t),t),t),t),t)-6*diff(diff(diff(x(t),t),t),t)+11*diff(diff(x(t),t),t)-6*diff(x(t),t) = exp(-3*t); 
dsolve(ode,x(t), singsol=all);
 

\[ x = c_1 \,{\mathrm e}^{t}+\frac {c_2 \,{\mathrm e}^{2 t}}{2}+\frac {c_3 \,{\mathrm e}^{3 t}}{3}+\frac {{\mathrm e}^{-3 t}}{360}+c_4 \]

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) = 6*diff( 
diff(_b(_a),_a),_a)-11*diff(_b(_a),_a)+6*_b(_a)+exp(-3*_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 t^{4}}x \left (t \right )-6 \frac {d^{3}}{d t^{3}}x \left (t \right )+11 \frac {d^{2}}{d t^{2}}x \left (t \right )-6 \frac {d}{d t}x \left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d^{4}}{d t^{4}}x \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{4}-6 r^{3}+11 r^{2}-6 r =0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left [0, 1, 2, 3\right ] \\ \bullet & {} & \textrm {Homogeneous solution from}\hspace {3pt} r =0 \\ {} & {} & x_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {Homogeneous solution from}\hspace {3pt} r =1 \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {Homogeneous solution from}\hspace {3pt} r =2 \\ {} & {} & x_{3}\left (t \right )={\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Homogeneous solution from}\hspace {3pt} r =3 \\ {} & {} & x_{4}\left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} x_{1}\left (t \right )+\mathit {C2} x_{2}\left (t \right )+\mathit {C3} x_{3}\left (t \right )+\mathit {C4} x_{4}\left (t \right )+x_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} +\mathit {C2} \,{\mathrm e}^{t}+\mathit {C3} \,{\mathrm e}^{2 t}+\mathit {C4} \,{\mathrm e}^{3 t}+x_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Define the forcing function of the ODE}\hspace {3pt} \\ {} & {} & f \left (t \right )={\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Form of the particular solution to the ODE where the}\hspace {3pt} u_{i}\left (t \right )\hspace {3pt}\textrm {are to be found}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (t \right ) x_{i}\left (t \right ) \\ {} & \circ & \textrm {Calculate the 1st derivative of}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & \frac {d}{d t}x_{p}\left (t \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d t}u_{i}\left (t \right )\right ) x_{i}\left (t \right )+u_{i}\left (t \right ) \left (\frac {d}{d t}x_{i}\left (t \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d t}u_{i}\left (t \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) x_{i}\left (t \right )=0 \\ {} & \circ & \textrm {Calculate the 2nd derivative of}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & \frac {d^{2}}{d t^{2}}x_{p}\left (t \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d}{d t}x_{i}\left (t \right )\right )+u_{i}\left (t \right ) \left (\frac {d^{2}}{d t^{2}}x_{i}\left (t \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d t}u_{i}\left (t \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d}{d t}x_{i}\left (t \right )\right )=0 \\ {} & \circ & \textrm {Calculate the 3rd derivative of}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & \frac {d^{3}}{d t^{3}}x_{p}\left (t \right )=\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{2}}{d t^{2}}x_{i}\left (t \right )\right )+u_{i}\left (t \right ) \left (\frac {d^{3}}{d t^{3}}x_{i}\left (t \right )\right )\right ) \\ {} & \circ & \textrm {Choose equation to add to a system of equations in}\hspace {3pt} \frac {d}{d t}u_{i}\left (t \right ) \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{2}}{d t^{2}}x_{i}\left (t \right )\right )=0 \\ {} & \circ & \textrm {The ODE is of the following form where the}\hspace {3pt} P_{i}\left (t \right )\hspace {3pt}\textrm {in this situation are the coefficients of the derivatives in the ODE}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{4}}x \left (t \right )+\left (\moverset {3}{\munderset {i =0}{\sum }}P_{i}\left (t \right ) \left (\frac {d^{i}}{d t^{i}}x \left (t \right )\right )\right )=f \left (t \right ) \\ {} & \circ & \textrm {Substitute}\hspace {3pt} x_{p}\left (t \right )=\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (t \right ) x_{i}\left (t \right )\hspace {3pt}\textrm {into the ODE}\hspace {3pt} \\ {} & {} & \left (\moverset {3}{\munderset {j =0}{\sum }}P_{j}\left (t \right ) \left (\moverset {4}{\munderset {i =1}{\sum }}u_{i}\left (t \right ) \left (\frac {d^{j}}{d t^{j}}x_{i}\left (t \right )\right )\right )\right )+\moverset {4}{\munderset {i =1}{\sum }}\left (\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{3}}{d t^{3}}x_{i}\left (t \right )\right )+u_{i}\left (t \right ) \left (\frac {d^{4}}{d t^{4}}x_{i}\left (t \right )\right )\right )=f \left (t \right ) \\ {} & \circ & \textrm {Rearrange the ODE}\hspace {3pt} \\ {} & {} & \moverset {4}{\munderset {i =1}{\sum }}\left (u_{i}\left (t \right )\cdot \left (\left (\moverset {3}{\munderset {j =0}{\sum }}P_{j}\left (t \right ) \left (\frac {d^{j}}{d t^{j}}x_{i}\left (t \right )\right )\right )+\frac {d^{4}}{d t^{4}}x_{i}\left (t \right )\right )+\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{3}}{d t^{3}}x_{i}\left (t \right )\right )\right )=f \left (t \right ) \\ {} & \circ & \textrm {Notice that}\hspace {3pt} x_{i}\left (t \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 t}u_{i}\left (t \right )\right ) \left (\frac {d^{3}}{d t^{3}}x_{i}\left (t \right )\right )=f \left (t \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 t}u_{i}\left (t \right )) \\ {} & {} & \left [\moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) x_{i}\left (t \right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d}{d t}x_{i}\left (t \right )\right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{2}}{d t^{2}}x_{i}\left (t \right )\right )=0, \moverset {4}{\munderset {i =1}{\sum }}\left (\frac {d}{d t}u_{i}\left (t \right )\right ) \left (\frac {d^{3}}{d t^{3}}x_{i}\left (t \right )\right )=f \left (t \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} x_{1}\left (t \right ) & x_{2}\left (t \right ) & x_{3}\left (t \right ) & x_{4}\left (t \right ) \\ \frac {d}{d t}x_{1}\left (t \right ) & \frac {d}{d t}x_{2}\left (t \right ) & \frac {d}{d t}x_{3}\left (t \right ) & \frac {d}{d t}x_{4}\left (t \right ) \\ \frac {d^{2}}{d t^{2}}x_{1}\left (t \right ) & \frac {d^{2}}{d t^{2}}x_{2}\left (t \right ) & \frac {d^{2}}{d t^{2}}x_{3}\left (t \right ) & \frac {d^{2}}{d t^{2}}x_{4}\left (t \right ) \\ \frac {d^{3}}{d t^{3}}x_{1}\left (t \right ) & \frac {d^{3}}{d t^{3}}x_{2}\left (t \right ) & \frac {d^{3}}{d t^{3}}x_{3}\left (t \right ) & \frac {d^{3}}{d t^{3}}x_{4}\left (t \right ) \end {array}\right ]\cdot \left [\begin {array}{c} \frac {d}{d t}u_{1}\left (t \right ) \\ \frac {d}{d t}u_{2}\left (t \right ) \\ \frac {d}{d t}u_{3}\left (t \right ) \\ \frac {d}{d t}u_{4}\left (t \right ) \end {array}\right ]=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Solve for the varied parameters}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} u_{1}\left (t \right ) \\ u_{2}\left (t \right ) \\ u_{3}\left (t \right ) \\ u_{4}\left (t \right ) \end {array}\right ]=\int W^{-1}\cdot \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ f \left (t \right ) \end {array}\right ]d t \\ {} & \circ & \textrm {Substitute in the homogeneous solutions and forcing function and solve}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} u_{1}\left (t \right ) \\ u_{2}\left (t \right ) \\ u_{3}\left (t \right ) \\ u_{4}\left (t \right ) \end {array}\right ]=\left [\begin {array}{c} \frac {{\mathrm e}^{-3 t}}{18} \\ -\frac {{\mathrm e}^{-3 t}}{8 \,{\mathrm e}^{t}} \\ \frac {{\mathrm e}^{-3 t}}{10 \,{\mathrm e}^{2 t}} \\ -\frac {{\mathrm e}^{-3 t}}{36 \,{\mathrm e}^{3 t}} \end {array}\right ] \\ & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {{\mathrm e}^{-3 t}}{360} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} +\mathit {C2} \,{\mathrm e}^{t}+\mathit {C3} \,{\mathrm e}^{2 t}+\mathit {C4} \,{\mathrm e}^{3 t}+\frac {{\mathrm e}^{-3 t}}{360} \end {array} \]

✓ Mathematica. Time used: 0.033 (sec). Leaf size: 45

ode=D[x[t],{t,4}]-6*D[x[t],{t,3}]+11*D[x[t],{t,2}]-6*D[x[t],t]==Exp[-3*t]; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to \frac {e^{-3 t}}{360}+c_1 e^t+\frac {1}{2} c_2 e^{2 t}+\frac {1}{3} c_3 e^{3 t}+c_4 \end{align*}

✓ Sympy. Time used: 0.160 (sec). Leaf size: 31

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(-6*Derivative(x(t), t) + 11*Derivative(x(t), (t, 2)) - 6*Derivative(x(t), (t, 3)) + Derivative(x(t), (t, 4)) - exp(-3*t),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

\[ x{\left (t \right )} = C_{1} + C_{2} e^{t} + C_{3} e^{2 t} + C_{4} e^{3 t} + \frac {e^{- 3 t}}{360} \]