Internal problem ID [14659]
Internal file name [OUTPUT/14339_Wednesday_April_03_2024_02_17_22_PM_61469210/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: 41.
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]]
\[ \boxed {y^{\prime \prime \prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 3] \end {align*}
The characteristic equation is \[ \lambda ^{3}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -\frac {1}{2}-\frac {i \sqrt {3}}{2}\\ \lambda _3 &= -\frac {1}{2}+\frac {i \sqrt {3}}{2} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{t}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) t} c_{2} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{t}\\ y_2 &= {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) t}\\ y_3 &= {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) t} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{t}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) t} c_{2} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) t} c_{3} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{1} {\mathrm e}^{t}+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) t} c_{2} +\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) t} c_{3} \end {align*}
substituting \(y^{\prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = \frac {i \left (c_{2} -c_{3} \right ) \sqrt {3}}{2}+c_{1} -\frac {c_{2}}{2}-\frac {c_{3}}{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{t}+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{2} {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) t} c_{2} +\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{2} {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) t} c_{3} \end {align*}
substituting \(y^{\prime \prime } = 3\) and \(t = 0\) in the above gives \begin {align*} 3 = \frac {i \left (c_{3} -c_{2} \right ) \sqrt {3}}{2}+c_{1} -\frac {c_{2}}{2}-\frac {c_{3}}{2}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=\frac {\left (-\sqrt {3}+3 i\right ) \sqrt {3}}{6}\\ c_{3}&=-\frac {\sqrt {3}\, \left (3 i+\sqrt {3}\right )}{6} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = {\mathrm e}^{t}+\frac {i \sqrt {3}\, {\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) t}{2}}}{2}-\frac {{\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) t}{2}}}{2}-\frac {i {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}} \sqrt {3}}{2}-\frac {{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}}}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (i \sqrt {3}-1\right ) {\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) t}{2}}}{2}-\frac {i {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}} \sqrt {3}}{2}+{\mathrm e}^{t}-\frac {{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (i \sqrt {3}-1\right ) {\mathrm e}^{\frac {\left (i \sqrt {3}-1\right ) t}{2}}}{2}-\frac {i {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}} \sqrt {3}}{2}+{\mathrm e}^{t}-\frac {{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) t}{2}}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3\right ] \\ \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 )=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 )=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 \\ 1 & 0 & 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 \\ 1 & 0 & 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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}} \\ 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}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right ) t}\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{-\frac {t}{2}}\cdot \left (\cos \left (\frac {\sqrt {3}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, t}{2}\right )\right )\cdot \left [\begin {array}{c} \frac {1}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {1}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} \frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{\left (-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}} \\ \frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{-\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ \cos \left (\frac {\sqrt {3}\, t}{2}\right )-\mathrm {I} \sin \left (\frac {\sqrt {3}\, t}{2}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}-\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ \frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \end {array}\right ]\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}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-\frac {t}{2}}\cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}-\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+\frac {\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \end {array}\right ]+{\mathrm e}^{-\frac {t}{2}} c_{3} \cdot \left [\begin {array}{c} -\frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ \frac {\cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{2}+\frac {\sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2} \\ -\sin \left (\frac {\sqrt {3}\, t}{2}\right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-\frac {t}{2}} \left (c_{3} \sqrt {3}+c_{2} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{2}-\frac {{\mathrm e}^{-\frac {t}{2}} \left (\sqrt {3}\, c_{2} -c_{3} \right ) \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{2}+c_{1} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{3} \sqrt {3}}{2}-\frac {c_{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{-\frac {t}{2}} \left (c_{3} \sqrt {3}+c_{2} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {t}{2}} \left (c_{3} \sqrt {3}+c_{2} \right ) \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {t}{2}} \left (\sqrt {3}\, c_{2} -c_{3} \right ) \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{4}-\frac {{\mathrm e}^{-\frac {t}{2}} \left (\sqrt {3}\, c_{2} -c_{3} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{4}+c_{1} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}-\frac {\left (\sqrt {3}\, c_{2} -c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {{\mathrm e}^{-\frac {t}{2}} \left (c_{3} \sqrt {3}+c_{2} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right )}{4}-\frac {{\mathrm e}^{-\frac {t}{2}} \left (c_{3} \sqrt {3}+c_{2} \right ) \sqrt {3}\, \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {t}{2}} \left (\sqrt {3}\, c_{2} -c_{3} \right ) \sin \left (\frac {\sqrt {3}\, t}{2}\right )}{4}+\frac {{\mathrm e}^{-\frac {t}{2}} \left (\sqrt {3}\, c_{2} -c_{3} \right ) \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \sqrt {3}}{4}+c_{1} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=\frac {c_{3} \sqrt {3}}{4}+\frac {c_{2}}{4}+\frac {\left (\sqrt {3}\, c_{2} -c_{3} \right ) \sqrt {3}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =1, c_{2} =2, c_{3} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{-\frac {t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right )-\sqrt {3}\, {\mathrm e}^{-\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )+{\mathrm e}^{t} \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: 0.031 (sec). Leaf size: 36
dsolve([diff(y(t),t$3)-y(t)=0,y(0) = 0, D(y)(0) = 0, (D@@2)(y)(0) = 3],y(t), singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{t}-\sqrt {3}\, {\mathrm e}^{-\frac {t}{2}} \sin \left (\frac {\sqrt {3}\, t}{2}\right )-{\mathrm e}^{-\frac {t}{2}} \cos \left (\frac {\sqrt {3}\, t}{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 52
DSolve[{y'''[t]-y[t]==0,{y[0]==0,y'[0]==0,y''[0]==3}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-t/2} \left (e^{3 t/2}-\sqrt {3} \sin \left (\frac {\sqrt {3} t}{2}\right )-\cos \left (\frac {\sqrt {3} t}{2}\right )\right ) \]