Internal problem ID [14643]
Internal file name [OUTPUT/14323_Wednesday_April_03_2024_02_17_18_PM_54653337/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 25.
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]]
\[ \boxed {y^{\prime \prime \prime }-5 y^{\prime }+2 y=0} \] The characteristic equation is \[ \lambda ^{3}-5 \lambda +2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= \sqrt {2}-1\\ \lambda _3 &= -1-\sqrt {2} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{2 t}+{\mathrm e}^{\left (\sqrt {2}-1\right ) t} c_{2} +{\mathrm e}^{\left (-1-\sqrt {2}\right ) t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{2 t}\\ y_2 &= {\mathrm e}^{\left (\sqrt {2}-1\right ) t}\\ y_3 &= {\mathrm e}^{\left (-1-\sqrt {2}\right ) t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{2 t}+{\mathrm e}^{\left (\sqrt {2}-1\right ) t} c_{2} +{\mathrm e}^{\left (-1-\sqrt {2}\right ) t} c_{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{2 t}+{\mathrm e}^{\left (\sqrt {2}-1\right ) t} c_{2} +{\mathrm e}^{\left (-1-\sqrt {2}\right ) t} c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-5 y^{\prime }+2 y=0 \\ \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 (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )=5 y_{2}\left (t \right )-2 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{3}^{\prime }\left (t \right )=5 y_{2}\left (t \right )-2 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & 5 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & 5 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\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 [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1-\sqrt {2}, \left [\begin {array}{c} \frac {1}{\left (-1-\sqrt {2}\right )^{2}} \\ \frac {1}{-1-\sqrt {2}} \\ 1 \end {array}\right ]\right ], \left [\sqrt {2}-1, \left [\begin {array}{c} \frac {1}{\left (\sqrt {2}-1\right )^{2}} \\ \frac {1}{\sqrt {2}-1} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1-\sqrt {2}, \left [\begin {array}{c} \frac {1}{\left (-1-\sqrt {2}\right )^{2}} \\ \frac {1}{-1-\sqrt {2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\left (-1-\sqrt {2}\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (-1-\sqrt {2}\right )^{2}} \\ \frac {1}{-1-\sqrt {2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\sqrt {2}-1, \left [\begin {array}{c} \frac {1}{\left (\sqrt {2}-1\right )^{2}} \\ \frac {1}{\sqrt {2}-1} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\left (\sqrt {2}-1\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (\sqrt {2}-1\right )^{2}} \\ \frac {1}{\sqrt {2}-1} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{\left (-1-\sqrt {2}\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (-1-\sqrt {2}\right )^{2}} \\ \frac {1}{-1-\sqrt {2}} \\ 1 \end {array}\right ]+c_{3} {\mathrm e}^{\left (\sqrt {2}-1\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (\sqrt {2}-1\right )^{2}} \\ \frac {1}{\sqrt {2}-1} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=c_{2} \left (3-2 \sqrt {2}\right ) {\mathrm e}^{-\left (1+\sqrt {2}\right ) t}+c_{3} \left (3+2 \sqrt {2}\right ) {\mathrm e}^{\left (\sqrt {2}-1\right ) t}+\frac {c_{1} {\mathrm e}^{2 t}}{4} \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: 32
dsolve(diff(y(t),t$3)-5*diff(y(t),t)+2*y(t)=0,y(t), singsol=all)
\[ y \left (t \right ) = c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{\left (\sqrt {2}-1\right ) t}+c_{3} {\mathrm e}^{-\left (1+\sqrt {2}\right ) t} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 43
DSolve[y'''[t]-5*y'[t]+2*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to c_1 e^{-\left (\left (1+\sqrt {2}\right ) t\right )}+c_2 e^{\left (\sqrt {2}-1\right ) t}+c_3 e^{2 t} \]