Internal problem ID [12765]
Internal file name [OUTPUT/11418_Friday_November_03_2023_06_32_47_AM_53036301/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.3, page 210
Problem number: 7.
ODE order: 4.
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
[[_high_order, _missing_x]]
\[ \boxed {36 y^{\prime \prime \prime \prime }-12 y^{\prime \prime \prime }-11 y^{\prime \prime }+2 y^{\prime }+y=0} \] The characteristic equation is \[ 36 \lambda ^{4}-12 \lambda ^{3}-11 \lambda ^{2}+2 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {1}{3}}\\ \lambda _2 &= -{\frac {1}{3}}\\ \lambda _3 &= {\frac {1}{2}}\\ \lambda _4 &= {\frac {1}{2}} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-\frac {x}{3}} c_{1} +x \,{\mathrm e}^{-\frac {x}{3}} c_{2} +{\mathrm e}^{\frac {x}{2}} c_{3} +x \,{\mathrm e}^{\frac {x}{2}} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-\frac {x}{3}}\\ y_2 &= x \,{\mathrm e}^{-\frac {x}{3}}\\ y_3 &= {\mathrm e}^{\frac {x}{2}}\\ y_4 &= x \,{\mathrm e}^{\frac {x}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\frac {x}{3}} c_{1} +x \,{\mathrm e}^{-\frac {x}{3}} c_{2} +{\mathrm e}^{\frac {x}{2}} c_{3} +x \,{\mathrm e}^{\frac {x}{2}} c_{4} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-\frac {x}{3}} c_{1} +x \,{\mathrm e}^{-\frac {x}{3}} c_{2} +{\mathrm e}^{\frac {x}{2}} c_{3} +x \,{\mathrm e}^{\frac {x}{2}} c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 36 y^{\prime \prime \prime \prime }-12 y^{\prime \prime \prime }-11 y^{\prime \prime }+2 y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\frac {y^{\prime \prime \prime }}{3}+\frac {11 y^{\prime \prime }}{36}-\frac {y^{\prime }}{18}-\frac {y}{36} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }-\frac {y^{\prime \prime \prime }}{3}-\frac {11 y^{\prime \prime }}{36}+\frac {y^{\prime }}{18}+\frac {y}{36}=0 \\ \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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=\frac {y_{4}\left (x \right )}{3}+\frac {11 y_{3}\left (x \right )}{36}-\frac {y_{2}\left (x \right )}{18}-\frac {y_{1}\left (x \right )}{36} \\ & {} & \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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=\frac {y_{4}\left (x \right )}{3}+\frac {11 y_{3}\left (x \right )}{36}-\frac {y_{2}\left (x \right )}{18}-\frac {y_{1}\left (x \right )}{36}\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 ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac {1}{36} & -\frac {1}{18} & \frac {11}{36} & \frac {1}{3} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac {1}{36} & -\frac {1}{18} & \frac {11}{36} & \frac {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 ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-\frac {1}{3}, \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{3}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{3}, \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -\frac {1}{3} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-\frac {x}{3}}\cdot \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =-\frac {1}{3}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =-\frac {1}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} -\frac {1}{3} \\ {} & {} & \left (\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac {1}{36} & -\frac {1}{18} & \frac {11}{36} & \frac {1}{3} \end {array}\right ]--\frac {1}{3}\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -81 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -\frac {1}{3} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-\frac {x}{3}}\cdot \left (x \cdot \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -81 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =\frac {1}{2}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{4}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{4}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac {1}{36} & -\frac {1}{18} & \frac {11}{36} & \frac {1}{3} \end {array}\right ]-\frac {1}{2}\cdot \left [\begin {array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -16 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} \frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}\left (x \right )={\mathrm e}^{\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -16 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}={\mathrm e}^{-\frac {x}{3}} c_{1} \cdot \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ]+{\mathrm e}^{-\frac {x}{3}} c_{2} \cdot \left (x \cdot \left [\begin {array}{c} -27 \\ 9 \\ -3 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -81 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right )+{\mathrm e}^{\frac {x}{2}} c_{3} \cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]+{\mathrm e}^{\frac {x}{2}} c_{4} \cdot \left (x \cdot \left [\begin {array}{c} 8 \\ 4 \\ 2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} -16 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-27 \left (\frac {8 \left (\left (2-x \right ) c_{4} -c_{3} \right ) {\mathrm e}^{\frac {5 x}{6}}}{27}+c_{2} \left (3+x \right )+c_{1} \right ) {\mathrm e}^{-\frac {x}{3}} \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 24
dsolve(36*diff(y(x),x$4)-12*diff(y(x),x$3)-11*diff(y(x),x$2)+2*diff(y(x),x)+y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{\frac {5 x}{6}}+c_{2} x +c_{1} \right ) {\mathrm e}^{-\frac {x}{3}} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 41
DSolve[36*y''''[x]-12*y'''[x]-11*y''[x]+2*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x/3} \left (c_3 e^{5 x/6}+x \left (c_4 e^{5 x/6}+c_2\right )+c_1\right ) \]