Internal problem ID [14671]
Internal file name [OUTPUT/14351_Wednesday_April_03_2024_02_17_26_PM_82600808/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: 63 (a).
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]]
\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }+2 y^{\prime }+6 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1, y^{\prime \prime }\left (0\right ) = -1] \end {align*}
The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}+2 \lambda +6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -3\\ \lambda _2 &= i \sqrt {2}\\ \lambda _3 &= -i \sqrt {2} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-3 t}+{\mathrm e}^{i \sqrt {2}\, t} c_{2} +{\mathrm e}^{-i \sqrt {2}\, t} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-3 t}\\ y_2 &= {\mathrm e}^{i \sqrt {2}\, t}\\ y_3 &= {\mathrm e}^{-i \sqrt {2}\, 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}^{-3 t}+{\mathrm e}^{i \sqrt {2}\, t} c_{2} +{\mathrm e}^{-i \sqrt {2}\, 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 } = -3 c_{1} {\mathrm e}^{-3 t}+i \sqrt {2}\, {\mathrm e}^{i \sqrt {2}\, t} c_{2} -i \sqrt {2}\, {\mathrm e}^{-i \sqrt {2}\, t} c_{3} \end {align*}
substituting \(y^{\prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = i \left (c_{2} -c_{3} \right ) \sqrt {2}-3 c_{1}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 9 c_{1} {\mathrm e}^{-3 t}-2 \,{\mathrm e}^{i \sqrt {2}\, t} c_{2} -2 \,{\mathrm e}^{-i \sqrt {2}\, t} c_{3} \end {align*}
substituting \(y^{\prime \prime } = -1\) and \(t = 0\) in the above gives \begin {align*} -1 = 9 c_{1} -2 c_{2} -2 c_{3}\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}&=-{\frac {1}{11}}\\ c_{2}&=\frac {\left (\sqrt {2}-8 i\right ) \sqrt {2}}{44}\\ c_{3}&=\frac {\sqrt {2}\, \left (\sqrt {2}+8 i\right )}{44} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {{\mathrm e}^{-3 t}}{11}+\frac {\cos \left (\sqrt {2}\, t \right )}{11}+\frac {4 \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{11} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-3 t}}{11}+\frac {\cos \left (\sqrt {2}\, t \right )}{11}+\frac {4 \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{11} \\ \end{align*}
Verification of solutions
\[ y = -\frac {{\mathrm e}^{-3 t}}{11}+\frac {\cos \left (\sqrt {2}\, t \right )}{11}+\frac {4 \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{11} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }+3 y^{\prime \prime }+2 y^{\prime }+6 y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-1\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 )=-3 y_{3}\left (t \right )-2 y_{2}\left (t \right )-6 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 )=-3 y_{3}\left (t \right )-2 y_{2}\left (t \right )-6 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 \\ -6 & -2 & -3 \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 \\ -6 & -2 & -3 \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 [-3, \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ -\frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ]\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}}_{1}={\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} \sqrt {2}\, t}\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {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 \\ {} & {} & \left (\cos \left (\sqrt {2}\, t \right )-\mathrm {I} \sin \left (\sqrt {2}\, t \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, t \right )}{2}+\frac {\mathrm {I} \sin \left (\sqrt {2}\, t \right )}{2} \\ \frac {\mathrm {I}}{2} \left (\cos \left (\sqrt {2}\, t \right )-\mathrm {I} \sin \left (\sqrt {2}\, t \right )\right ) \sqrt {2} \\ \cos \left (\sqrt {2}\, t \right )-\mathrm {I} \sin \left (\sqrt {2}\, t \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 )=\left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, t \right )}{2} \\ \frac {\sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{2} \\ \cos \left (\sqrt {2}\, t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )=\left [\begin {array}{c} \frac {\sin \left (\sqrt {2}\, t \right )}{2} \\ \frac {\cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{2} \\ -\sin \left (\sqrt {2}\, t \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}^{-3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {c_{3} \sin \left (\sqrt {2}\, t \right )}{2}-\frac {c_{2} \cos \left (\sqrt {2}\, t \right )}{2} \\ \frac {c_{3} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{2}+\frac {c_{2} \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{2} \\ -c_{3} \sin \left (\sqrt {2}\, t \right )+c_{2} \cos \left (\sqrt {2}\, t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{1} {\mathrm e}^{-3 t}}{9}+\frac {c_{3} \sin \left (\sqrt {2}\, t \right )}{2}-\frac {c_{2} \cos \left (\sqrt {2}\, t \right )}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\frac {c_{1}}{9}-\frac {c_{2}}{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{-3 t}}{3}+\frac {c_{3} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{2}+\frac {c_{2} \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-\frac {c_{1}}{3}+\frac {c_{3} \sqrt {2}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{1} {\mathrm e}^{-3 t}-c_{3} \sin \left (\sqrt {2}\, t \right )+c_{2} \cos \left (\sqrt {2}\, t \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=c_{1} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-\frac {9}{11}, c_{2} =-\frac {2}{11}, c_{3} =\frac {8 \sqrt {2}}{11}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-3 t}}{11}+\frac {\cos \left (\sqrt {2}\, t \right )}{11}+\frac {4 \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{11} \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: 30
dsolve([diff(y(t),t$3)+3*diff(y(t),t$2)+2*diff(y(t),t)+6*y(t)=0,y(0) = 0, D(y)(0) = 1, (D@@2)(y)(0) = -1],y(t), singsol=all)
\[ y \left (t \right ) = -\frac {{\mathrm e}^{-3 t}}{11}+\frac {4 \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{11}+\frac {\cos \left (\sqrt {2}\, t \right )}{11} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 40
DSolve[{y'''[t]+3*y''[t]+2*y'[t]+6*y[t]==0,{y[0]==0,y'[0]==1,y''[0]==-1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{11} \left (-e^{-3 t}+4 \sqrt {2} \sin \left (\sqrt {2} t\right )+\cos \left (\sqrt {2} t\right )\right ) \]