Internal problem ID [12307]
Internal file name [OUTPUT/10960_Saturday_September_30_2023_08_26_37_PM_5883988/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 27.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_laplace"
Maple gives the following as the ode type
[[_high_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime \prime }+13 y^{\prime \prime }+36 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = -1, y^{\prime \prime }\left (0\right ) = 5, y^{\prime \prime \prime }\left (0\right ) = 19] \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 )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}
The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )+13 s^{2} Y \left (s \right )-13 y^{\prime }\left (0\right )-13 s y \left (0\right )+36 Y \left (s \right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=-1\\ y^{\prime \prime }\left (0\right )&=5\\ y^{\prime \prime \prime }\left (0\right )&=19 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-6-5 s +s^{2}+13 s^{2} Y \left (s \right )+36 Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = -\frac {s^{2}-5 s -6}{s^{4}+13 s^{2}+36} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {\frac {1}{2}-\frac {i}{2}}{s -2 i}+\frac {\frac {1}{2}+\frac {i}{2}}{s +2 i}+\frac {-\frac {1}{2}+\frac {i}{2}}{s -3 i}+\frac {-\frac {1}{2}-\frac {i}{2}}{s +3 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\frac {1}{2}-\frac {i}{2}}{s -2 i}\right ) &= \left (\frac {1}{2}-\frac {i}{2}\right ) {\mathrm e}^{2 i t}\\ \mathcal {L}^{-1}\left (\frac {\frac {1}{2}+\frac {i}{2}}{s +2 i}\right ) &= \left (\frac {1}{2}+\frac {i}{2}\right ) {\mathrm e}^{-2 i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}+\frac {i}{2}}{s -3 i}\right ) &= \left (-\frac {1}{2}+\frac {i}{2}\right ) {\mathrm e}^{3 i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{2}-\frac {i}{2}}{s +3 i}\right ) &= \left (-\frac {1}{2}-\frac {i}{2}\right ) {\mathrm e}^{-3 i t} \end {align*}
Adding the above results and simplifying gives \[ y=-\cos \left (3 t \right )-\sin \left (3 t \right )+\cos \left (2 t \right )+\sin \left (2 t \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\cos \left (3 t \right )-\sin \left (3 t \right )+\cos \left (2 t \right )+\sin \left (2 t \right ) \\ \end{align*}
Verification of solutions
\[ y = -\cos \left (3 t \right )-\sin \left (3 t \right )+\cos \left (2 t \right )+\sin \left (2 t \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }+13 y^{\prime \prime }+36 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 \}}}}=5, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=19\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \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 {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (t \right )=-13 y_{3}\left (t \right )-36 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_{4}\left (t \right )=y_{3}^{\prime }\left (t \right ), y_{4}^{\prime }\left (t \right )=-13 y_{3}\left (t \right )-36 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 ) \\ y_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -36 & 0 & -13 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -36 & 0 & -13 & 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 [-3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [3 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ -\frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-3 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 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 (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ -\frac {\cos \left (3 t \right )}{9}+\frac {\mathrm {I} \sin \left (3 t \right )}{9} \\ \frac {\mathrm {I}}{3} \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{1}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (3 t \right )}{27} \\ -\frac {\cos \left (3 t \right )}{9} \\ \frac {\sin \left (3 t \right )}{3} \\ \cos \left (3 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{2}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (3 t \right )}{27} \\ \frac {\sin \left (3 t \right )}{9} \\ \frac {\cos \left (3 t \right )}{3} \\ -\sin \left (3 t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{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 (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ -\frac {\cos \left (2 t \right )}{4}+\frac {\mathrm {I} \sin \left (2 t \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (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}}_{3}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (2 t \right )}{8} \\ -\frac {\cos \left (2 t \right )}{4} \\ \frac {\sin \left (2 t \right )}{2} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{8} \\ \frac {\sin \left (2 t \right )}{4} \\ \frac {\cos \left (2 t \right )}{2} \\ -\sin \left (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}\left (t \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} -\frac {c_{4} \cos \left (2 t \right )}{8}-\frac {c_{3} \sin \left (2 t \right )}{8}-\frac {c_{2} \cos \left (3 t \right )}{27}-\frac {c_{1} \sin \left (3 t \right )}{27} \\ \frac {c_{4} \sin \left (2 t \right )}{4}-\frac {c_{3} \cos \left (2 t \right )}{4}+\frac {c_{2} \sin \left (3 t \right )}{9}-\frac {c_{1} \cos \left (3 t \right )}{9} \\ \frac {c_{4} \cos \left (2 t \right )}{2}+\frac {c_{3} \sin \left (2 t \right )}{2}+\frac {c_{2} \cos \left (3 t \right )}{3}+\frac {c_{1} \sin \left (3 t \right )}{3} \\ -c_{4} \sin \left (2 t \right )+c_{3} \cos \left (2 t \right )-c_{2} \sin \left (3 t \right )+c_{1} \cos \left (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {c_{4} \cos \left (2 t \right )}{8}-\frac {c_{3} \sin \left (2 t \right )}{8}-\frac {c_{2} \cos \left (3 t \right )}{27}-\frac {c_{1} \sin \left (3 t \right )}{27} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{4}}{8}-\frac {c_{2}}{27} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{4} \sin \left (2 t \right )}{4}-\frac {c_{3} \cos \left (2 t \right )}{4}+\frac {c_{2} \sin \left (3 t \right )}{9}-\frac {c_{1} \cos \left (3 t \right )}{9} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-1 \\ {} & {} & -1=-\frac {c_{3}}{4}-\frac {c_{1}}{9} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {c_{4} \cos \left (2 t \right )}{2}+\frac {c_{3} \sin \left (2 t \right )}{2}+\frac {c_{2} \cos \left (3 t \right )}{3}+\frac {c_{1} \sin \left (3 t \right )}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=\frac {c_{4}}{2}+\frac {c_{2}}{3} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-c_{4} \sin \left (2 t \right )+c_{3} \cos \left (2 t \right )-c_{2} \sin \left (3 t \right )+c_{1} \cos \left (3 t \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=19 \\ {} & {} & 19=c_{3} +c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =27, c_{2} =27, c_{3} =-8, c_{4} =-8\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\cos \left (3 t \right )-\sin \left (3 t \right )+\cos \left (2 t \right )+\sin \left (2 t \right ) \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 4.953 (sec). Leaf size: 25
dsolve([diff(y(t),t$4)+13*diff(y(t),t$2)+36*y(t)=0,y(0) = 0, D(y)(0) = -1, (D@@2)(y)(0) = 5, (D@@3)(y)(0) = 19],y(t), singsol=all)
\[ y \left (t \right ) = \cos \left (2 t \right )+\sin \left (2 t \right )-\cos \left (3 t \right )-\sin \left (3 t \right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 26
DSolve[{y''''[t]+13*y''[t]+36*y[t]==0,{y[0]==0,y'[0]==-1,y''[0]==5,y'''[0]==19}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \sin (2 t)-\sin (3 t)+\cos (2 t)-\cos (3 t) \]