Internal problem ID [12304]
Internal file name [OUTPUT/10957_Saturday_September_30_2023_08_26_36_PM_73489955/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.5 Laplace transform. Homogeneous equations. Problems page 357
Problem number: Problem 24.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_3rd_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime }+4 y^{\prime \prime }+29 y^{\prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 5, y^{\prime \prime }\left (0\right ) = -20] \end {align*}
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )+4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+29 s Y \left (s \right )-29 y \left (0\right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y^{\prime }\left (0\right )&=5\\ y^{\prime \prime }\left (0\right )&=-20 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )-29-9 s -s^{2}+4 s^{2} Y \left (s \right )+29 s Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s^{2}+9 s +29}{s \left (s^{2}+4 s +29\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{s}-\frac {i}{2 \left (s +2-5 i\right )}+\frac {i}{2 s +4+10 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{s}\right ) &= 1\\ \mathcal {L}^{-1}\left (-\frac {i}{2 \left (s +2-5 i\right )}\right ) &= -\frac {i {\mathrm e}^{\left (-2+5 i\right ) t}}{2}\\ \mathcal {L}^{-1}\left (\frac {i}{2 s +4+10 i}\right ) &= \frac {i {\mathrm e}^{\left (-2-5 i\right ) t}}{2} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{-2 t} \sin \left (5 t \right )+1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 t} \sin \left (5 t \right )+1 \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-2 t} \sin \left (5 t \right )+1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }+4 y^{\prime \prime }+29 y^{\prime }=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-20\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 )=-4 y_{3}\left (t \right )-29 y_{2}\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 )=-4 y_{3}\left (t \right )-29 y_{2}\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 \\ 0 & -29 & -4 \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 \\ 0 & -29 & -4 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [-2-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ], \left [-2+5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}+\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}-\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-5 \,\mathrm {I}, \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-5 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 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}^{-2 t}\cdot \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right )\cdot \left [\begin {array}{c} -\frac {21}{841}-\frac {20 \,\mathrm {I}}{841} \\ -\frac {2}{29}+\frac {5 \,\mathrm {I}}{29} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \left (-\frac {21}{841}-\frac {20 \,\mathrm {I}}{841}\right ) \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right ) \\ \left (-\frac {2}{29}+\frac {5 \,\mathrm {I}}{29}\right ) \left (\cos \left (5 t \right )-\mathrm {I} \sin \left (5 t \right )\right ) \\ \cos \left (5 t \right )-\mathrm {I} \sin \left (5 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 )={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {21 \cos \left (5 t \right )}{841}-\frac {20 \sin \left (5 t \right )}{841} \\ -\frac {2 \cos \left (5 t \right )}{29}+\frac {5 \sin \left (5 t \right )}{29} \\ \cos \left (5 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {21 \sin \left (5 t \right )}{841}-\frac {20 \cos \left (5 t \right )}{841} \\ \frac {2 \sin \left (5 t \right )}{29}+\frac {5 \cos \left (5 t \right )}{29} \\ -\sin \left (5 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_{2} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {21 \cos \left (5 t \right )}{841}-\frac {20 \sin \left (5 t \right )}{841} \\ -\frac {2 \cos \left (5 t \right )}{29}+\frac {5 \sin \left (5 t \right )}{29} \\ \cos \left (5 t \right ) \end {array}\right ]+c_{3} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {21 \sin \left (5 t \right )}{841}-\frac {20 \cos \left (5 t \right )}{841} \\ \frac {2 \sin \left (5 t \right )}{29}+\frac {5 \cos \left (5 t \right )}{29} \\ -\sin \left (5 t \right ) \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (-21 c_{2} -20 c_{3} \right ) \cos \left (5 t \right )-20 \sin \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841}+c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {21 c_{2}}{841}-\frac {20 c_{3}}{841}+c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (-5 \left (-21 c_{2} -20 c_{3} \right ) \sin \left (5 t \right )-100 \cos \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841}-\frac {2 \left (\left (-21 c_{2} -20 c_{3} \right ) \cos \left (5 t \right )-20 \sin \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=-\frac {2 c_{2}}{29}+\frac {5 c_{3}}{29} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (-25 \left (-21 c_{2} -20 c_{3} \right ) \cos \left (5 t \right )+500 \sin \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841}-\frac {4 \left (-5 \left (-21 c_{2} -20 c_{3} \right ) \sin \left (5 t \right )-100 \cos \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841}+\frac {4 \left (\left (-21 c_{2} -20 c_{3} \right ) \cos \left (5 t \right )-20 \sin \left (5 t \right ) \left (c_{2} -\frac {21 c_{3}}{20}\right )\right ) {\mathrm e}^{-2 t}}{841} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-20 \\ {} & {} & -20=c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =1, c_{2} =-20, c_{3} =21\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-2 t} \sin \left (5 t \right )+1 \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: 4.625 (sec). Leaf size: 15
dsolve([diff(y(t),t$3)+4*diff(y(t),t$2)+29*diff(y(t),t)=0,y(0) = 1, D(y)(0) = 5, (D@@2)(y)(0) = -20],y(t), singsol=all)
\[ y \left (t \right ) = 1+{\mathrm e}^{-2 t} \sin \left (5 t \right ) \]
✓ Solution by Mathematica
Time used: 0.58 (sec). Leaf size: 49
DSolve[{y'''[t]+4*y''[t]-20*y'[t]==0,{y[0]==1,y'[0]==5,y''[0]==-20}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {5 e^{2 \left (\sqrt {6}-1\right ) t}}{4 \sqrt {6}}-\frac {5 e^{-2 \left (1+\sqrt {6}\right ) t}}{4 \sqrt {6}}+1 \]