12.26 problem 26

Internal problem ID [11853]
Internal file name [OUTPUT/11863_Saturday_April_13_2024_01_13_10_AM_74535894/index.tex]

Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section: Chapter 4, Section 4.4. Variation of parameters. Exercises page 162
Problem number: 26.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"

Maple gives the following as the ode type

[[_3rd_order, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }-y^{\prime }+3 y=x^{2} {\mathrm e}^{x}} \] 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 \[ y^{\prime \prime \prime }-3 y^{\prime \prime }-y^{\prime }+3 y = 0 \] The characteristic equation is \[ \lambda ^{3}-3 \lambda ^{2}-\lambda +3 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 3\\ \lambda _3 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-3 y^{\prime \prime }-y^{\prime }+3 y = x^{2} {\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x^{2} {\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-x}, {\mathrm e}^{3 x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}, x^{3} {\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} x^{3} {\mathrm e}^{x}+A_{3} x^{2} {\mathrm e}^{x} \] 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 \[ -8 A_{3} x \,{\mathrm e}^{x}-12 A_{2} x^{2} {\mathrm e}^{x}-4 A_{1} {\mathrm e}^{x}+6 A_{2} {\mathrm e}^{x} = x^{2} {\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{8}}, A_{2} = -{\frac {1}{12}}, A_{3} = 0\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{x}}{8}-\frac {x^{3} {\mathrm e}^{x}}{12} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3}\right ) + \left (-\frac {x \,{\mathrm e}^{x}}{8}-\frac {x^{3} {\mathrm e}^{x}}{12}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3} -\frac {x \,{\mathrm e}^{x}}{8}-\frac {x^{3} {\mathrm e}^{x}}{12} \\ \end{align*}

Verification of solutions

\[ y = c_{1} {\mathrm e}^{-x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3} -\frac {x \,{\mathrm e}^{x}}{8}-\frac {x^{3} {\mathrm e}^{x}}{12} \] Verified OK.

12.26.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-3 y^{\prime \prime }-y^{\prime }+3 y=x^{2} {\mathrm e}^{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \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 \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=x^{2} {\mathrm e}^{x}+3 y_{3}\left (x \right )+y_{2}\left (x \right )-3 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=x^{2} {\mathrm e}^{x}+3 y_{3}\left (x \right )+y_{2}\left (x \right )-3 y_{1}\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} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -3 & 1 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ x^{2} {\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ x^{2} {\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -3 & 1 & 3 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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 [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}}_{2}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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 )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\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} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{x} & {\mathrm e}^{3 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 \frac {1}{\phi \left (0\right )} \\ {} & \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} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{9} \\ -{\mathrm e}^{-x} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{x} & {\mathrm e}^{3 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & \frac {1}{9} \\ -1 & 1 & \frac {1}{3} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {3 \,{\mathrm e}^{-x}}{8}+\frac {3 \,{\mathrm e}^{x}}{4}-\frac {{\mathrm e}^{3 x}}{8} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {{\mathrm e}^{-x}}{8}-\frac {{\mathrm e}^{x}}{4}+\frac {{\mathrm e}^{3 x}}{8} \\ -\frac {3 \,{\mathrm e}^{-x}}{8}+\frac {3 \,{\mathrm e}^{x}}{4}-\frac {3 \,{\mathrm e}^{3 x}}{8} & \frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & -\frac {{\mathrm e}^{-x}}{8}-\frac {{\mathrm e}^{x}}{4}+\frac {3 \,{\mathrm e}^{3 x}}{8} \\ \frac {3 \,{\mathrm e}^{-x}}{8}+\frac {3 \,{\mathrm e}^{x}}{4}-\frac {9 \,{\mathrm e}^{3 x}}{8} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {{\mathrm e}^{-x}}{8}-\frac {{\mathrm e}^{x}}{4}+\frac {9 \,{\mathrm e}^{3 x}}{8} \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} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\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}\frac {1}{\Phi \left (s \right )}\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}\frac {1}{\Phi \left (s \right )}\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 {{\mathrm e}^{-x}}{32}+\frac {{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-3 x \right ) {\mathrm e}^{x}}{24} \\ \frac {{\mathrm e}^{-x}}{32}+\frac {3 \,{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-6 x^{2}-3 x -3\right ) {\mathrm e}^{x}}{24} \\ -\frac {{\mathrm e}^{-x}}{32}+\frac {9 \,{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-12 x^{2}-15 x -6\right ) {\mathrm e}^{x}}{24} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -\frac {{\mathrm e}^{-x}}{32}+\frac {{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-3 x \right ) {\mathrm e}^{x}}{24} \\ \frac {{\mathrm e}^{-x}}{32}+\frac {3 \,{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-6 x^{2}-3 x -3\right ) {\mathrm e}^{x}}{24} \\ -\frac {{\mathrm e}^{-x}}{32}+\frac {9 \,{\mathrm e}^{3 x}}{32}+\frac {\left (-2 x^{3}-12 x^{2}-15 x -6\right ) {\mathrm e}^{x}}{24} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-1+32 c_{1} \right ) {\mathrm e}^{-x}}{32}+\frac {\left (32 c_{3} +9\right ) {\mathrm e}^{3 x}}{288}-\frac {\left (x^{3}+\frac {3}{2} x -12 c_{2} \right ) {\mathrm e}^{x}}{12} \end {array} \]

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`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 31

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

\[ y \left (x \right ) = c_{2} {\mathrm e}^{-x}+c_{3} {\mathrm e}^{3 x}-\frac {\left (x^{3}+\frac {3}{2} x -12 c_{1} \right ) {\mathrm e}^{x}}{12} \]

✓ Solution by Mathematica

Time used: 0.038 (sec). Leaf size: 41

DSolve[y'''[x]-3*y''[x]-y'[x]+3*y[x]==x^2*Exp[x],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^x \left (-\frac {x^3}{12}-\frac {x}{8}+c_2\right )+c_1 e^{-x}+c_3 e^{3 x} \]