12.15 problem 15

Internal problem ID [2838]
Book : Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section : Exercise 20, page 90
Problem number : 15
Date solved : Thursday, October 17, 2024 at 03:11:54 AM
CAS classification : [[_3rd_order, _missing_y]]

Solve

\begin{align*} y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }&={\mathrm e}^{2 x}+\sin \left (x \right ) \end{align*}

12.15.1 Solved as higher order constant coeff ode

Time used: 0.129 (sec)

The characteristic equation is

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

The roots of the above equation are

\begin{align*} \lambda _1 &= 0\\ \lambda _2 &= 4\\ \lambda _3 &= -1 \end{align*}

Therefore the homogeneous solution is

\[ y_h(x)={\mathrm e}^{-x} c_1 +c_2 +{\mathrm e}^{4 x} c_3 \]

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

\begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= 1\\ y_3 &= {\mathrm e}^{4 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

\[ y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime } = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime } = {\mathrm e}^{2 x}+\sin \left (x \right ) \]

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

Looking at the RHS of the ode, which is

\[ {\mathrm e}^{2 x}+\sin \left (x \right ) \]

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

\[ [\{{\mathrm e}^{2 x}\}, \{\cos \left (x \right ), \sin \left (x \right )\}] \]

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

\[ \{1, {\mathrm e}^{-x}, {\mathrm e}^{4 x}\} \]

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.

\[ y_p = A_{1} {\mathrm e}^{2 x}+A_{2} \cos \left (x \right )+A_{3} \sin \left (x \right ) \]

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

\[ -12 A_{1} {\mathrm e}^{2 x}+5 A_{2} \sin \left (x \right )-5 A_{3} \cos \left (x \right )+3 A_{2} \cos \left (x \right )+3 A_{3} \sin \left (x \right ) = {\mathrm e}^{2 x}+\sin \left (x \right ) \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = -{\frac {1}{12}}, A_{2} = {\frac {5}{34}}, A_{3} = {\frac {3}{34}}\right ] \]

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

\[ y_p = -\frac {{\mathrm e}^{2 x}}{12}+\frac {5 \cos \left (x \right )}{34}+\frac {3 \sin \left (x \right )}{34} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_1 +c_2 +{\mathrm e}^{4 x} c_3\right ) + \left (-\frac {{\mathrm e}^{2 x}}{12}+\frac {5 \cos \left (x \right )}{34}+\frac {3 \sin \left (x \right )}{34}\right ) \\ \end{align*}

12.15.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )-3 \frac {d^{2}}{d x^{2}}y \left (x \right )-4 \frac {d}{d x}y \left (x \right )={\mathrm e}^{2 x}+\sin \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right ) \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=\frac {d}{d x}y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}y_{3}\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y_{3}\left (x \right )={\mathrm e}^{2 x}+\sin \left (x \right )+3 y_{3}\left (x \right )+4 y_{2}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=\frac {d}{d x}y_{1}\left (x \right ), y_{3}\left (x \right )=\frac {d}{d x}y_{2}\left (x \right ), \frac {d}{d x}y_{3}\left (x \right )={\mathrm e}^{2 x}+\sin \left (x \right )+3 y_{3}\left (x \right )+4 y_{2}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 4 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{2 x}+\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{2 x}+\sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 4 & 3 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{4 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & \frac {{\mathrm e}^{4 x}}{16} \\ -{\mathrm e}^{-x} & 0 & \frac {{\mathrm e}^{4 x}}{4} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{4 x} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \phi \left (0\right )^{-1} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & \frac {{\mathrm e}^{4 x}}{16} \\ -{\mathrm e}^{-x} & 0 & \frac {{\mathrm e}^{4 x}}{4} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{4 x} \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & 1 & \frac {1}{16} \\ -1 & 0 & \frac {1}{4} \\ 1 & 0 & 1 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} 1 & -\frac {4 \,{\mathrm e}^{-x}}{5}+\frac {3}{4}+\frac {{\mathrm e}^{4 x}}{20} & \frac {{\mathrm e}^{-x}}{5}-\frac {1}{4}+\frac {{\mathrm e}^{4 x}}{20} \\ 0 & \frac {4 \,{\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4 x}}{5} & -\frac {{\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4 x}}{5} \\ 0 & -\frac {4 \,{\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{4 x}}{5} & \frac {{\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{4 x}}{5} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )=\Phi \left (x \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {3 \sin \left (x \right )}{34}+\frac {5 \cos \left (x \right )}{34}+\frac {19 \,{\mathrm e}^{4 x}}{680}-\frac {{\mathrm e}^{2 x}}{12}-\frac {1}{8}+\frac {{\mathrm e}^{-x}}{30} \\ \frac {19 \,{\mathrm e}^{4 x}}{170}-\frac {{\mathrm e}^{2 x}}{6}-\frac {5 \sin \left (x \right )}{34}+\frac {3 \cos \left (x \right )}{34}-\frac {{\mathrm e}^{-x}}{30} \\ \frac {38 \,{\mathrm e}^{4 x}}{85}-\frac {{\mathrm e}^{2 x}}{3}-\frac {3 \sin \left (x \right )}{34}-\frac {5 \cos \left (x \right )}{34}+\frac {{\mathrm e}^{-x}}{30} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} \frac {3 \sin \left (x \right )}{34}+\frac {5 \cos \left (x \right )}{34}+\frac {19 \,{\mathrm e}^{4 x}}{680}-\frac {{\mathrm e}^{2 x}}{12}-\frac {1}{8}+\frac {{\mathrm e}^{-x}}{30} \\ \frac {19 \,{\mathrm e}^{4 x}}{170}-\frac {{\mathrm e}^{2 x}}{6}-\frac {5 \sin \left (x \right )}{34}+\frac {3 \cos \left (x \right )}{34}-\frac {{\mathrm e}^{-x}}{30} \\ \frac {38 \,{\mathrm e}^{4 x}}{85}-\frac {{\mathrm e}^{2 x}}{3}-\frac {3 \sin \left (x \right )}{34}-\frac {5 \cos \left (x \right )}{34}+\frac {{\mathrm e}^{-x}}{30} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (4080 \mathit {C1} +136\right ) {\mathrm e}^{-x}}{4080}+\frac {\left (255 \mathit {C3} +114\right ) {\mathrm e}^{4 x}}{4080}+\mathit {C2} +\frac {5 \cos \left (x \right )}{34}+\frac {3 \sin \left (x \right )}{34}-\frac {{\mathrm e}^{2 x}}{12}-\frac {1}{8} \end {array} \]

12.15.3 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] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = 4*_b(_a)+3*(diff(_b(_a), _a))+exp(2*_a)+sin(_a), _b(_a)`   *** Sublev 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying high order exact linear fully integrable 
   trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
   trying a double symmetry of the form [xi=0, eta=F(x)] 
   -> Try solving first the homogeneous part of the ODE 
      checking if the LODE has constant coefficients 
      <- constant coefficients successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
 

12.15.4 Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 34

dsolve(-4*diff(y(x),x)-3*diff(diff(y(x),x),x)+diff(diff(diff(y(x),x),x),x) = exp(2*x)+sin(x), 
       y(x),singsol=all)
 

\[ y = -c_2 \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{4 x} c_{1}}{4}+\frac {3 \sin \left (x \right )}{34}-\frac {{\mathrm e}^{2 x}}{12}+\frac {5 \cos \left (x \right )}{34}+c_3 \]

12.15.5 Mathematica DSolve solution

Solving time : 0.257 (sec)
Leaf size : 49

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

\[ y(x)\to -\frac {e^{2 x}}{12}+\frac {3 \sin (x)}{34}+\frac {5 \cos (x)}{34}+c_1 \left (-e^{-x}\right )+\frac {1}{4} c_2 e^{4 x}+c_3 \]