Internal problem ID [5049]
Internal file name [OUTPUT/4542_Sunday_June_05_2022_03_00_35_PM_42275176/index.tex]
Book: Basic Training in Mathematics. By R. Shankar. Plenum Press. NY. 1995
Section: Chapter 10, Differential equations. Section 10.2, ODEs with constant Coefficients. page
307
Problem number: 10.2.8 part(3).
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, _missing_x]]
\[ \boxed {x^{\prime \prime \prime }-3 x^{\prime \prime }-9 x^{\prime }-5 x=0} \] The characteristic equation is \[ \lambda ^{3}-3 \lambda ^{2}-9 \lambda -5 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 5\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ x_h(t)={\mathrm e}^{-t} c_{1} +t \,{\mathrm e}^{-t} c_{2} +{\mathrm e}^{5 t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} x_1 &= {\mathrm e}^{-t}\\ x_2 &= t \,{\mathrm e}^{-t}\\ x_3 &= {\mathrm e}^{5 t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= {\mathrm e}^{-t} c_{1} +t \,{\mathrm e}^{-t} c_{2} +{\mathrm e}^{5 t} c_{3} \\ \end{align*}
Verification of solutions
\[ x = {\mathrm e}^{-t} c_{1} +t \,{\mathrm e}^{-t} c_{2} +{\mathrm e}^{5 t} c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime \prime }-3 x^{\prime \prime }-9 x^{\prime }-5 x=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & x^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{1}\left (t \right ) \\ {} & {} & x_{1}\left (t \right )=x \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{2}\left (t \right ) \\ {} & {} & x_{2}\left (t \right )=x^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} x_{3}\left (t \right ) \\ {} & {} & x_{3}\left (t \right )=x^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} x_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & x_{3}^{\prime }\left (t \right )=3 x_{3}\left (t \right )+9 x_{2}\left (t \right )+5 x_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [x_{2}\left (t \right )=x_{1}^{\prime }\left (t \right ), x_{3}\left (t \right )=x_{2}^{\prime }\left (t \right ), x_{3}^{\prime }\left (t \right )=3 x_{3}\left (t \right )+9 x_{2}\left (t \right )+5 x_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}\left (t \right )=\left [\begin {array}{c} x_{1}\left (t \right ) \\ x_{2}\left (t \right ) \\ x_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & 9 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{x}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & 9 & 3 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{x}}\left (t \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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [5, \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{x}}_{1}\left (t \right )={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 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 =-1\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{2}\left (t \right )={\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} t \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 =-1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{x}}_{2}\left (t \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda t} A \right )\cdot \left (t {\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 t} \left (t {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda t} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda t} \left (\lambda t {\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 }{x}}_{2}\left (t \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} -1 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & 9 & 3 \end {array}\right ]-\left (-1\right )\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -1 \\ {} & {} & {\moverset {\rightarrow }{x}}_{2}\left (t \right )={\mathrm e}^{-t}\cdot \left (t \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [5, \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}_{3}={\mathrm e}^{5 t}\cdot \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}=c_{1} {\moverset {\rightarrow }{x}}_{1}\left (t \right )+c_{2} {\moverset {\rightarrow }{x}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{x}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{x}}={\mathrm e}^{-t} c_{1} \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+{\mathrm e}^{-t} c_{2} \cdot \left (t \cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right )+{\mathrm e}^{5 t} c_{3} \cdot \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & x=\left (\left (t +1\right ) c_{2} +c_{1} \right ) {\mathrm e}^{-t}+\frac {{\mathrm e}^{5 t} c_{3}}{25} \end {array} \]
Maple trace
`Methods for third 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: 21
dsolve(diff(x(t),t$3)-3*diff(x(t),t$2)-9*diff(x(t),t)-5*x(t)=0,x(t), singsol=all)
\[ x \left (t \right ) = \left (c_{3} t +c_{2} \right ) {\mathrm e}^{-t}+c_{1} {\mathrm e}^{5 t} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 26
DSolve[x'''[t]-3*x''[t]-9*x'[t]-5*x[t]==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to e^{-t} \left (c_2 t+c_3 e^{6 t}+c_1\right ) \]