Internal problem ID [839]
Internal file name [OUTPUT/839_Sunday_June_05_2022_01_50_52_AM_19041891/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 6.2, The Laplace Transform. Solution of Initial Value Problems. page
255
Problem number: 14.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 1, y^{\prime \prime \prime }\left (0\right ) = 0] \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 \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-4 Y \left (s \right ) = 0\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=1\\ y^{\prime \prime \prime }\left (0\right )&=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-s -s^{3}-4 Y \left (s \right ) = 0 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s \left (s^{2}+1\right )}{s^{4}-4} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{8 \left (s -\sqrt {2}\right )}+\frac {3}{8 \left (s +\sqrt {2}\right )}+\frac {1}{8 s -8 i \sqrt {2}}+\frac {1}{8 s +8 i \sqrt {2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{8 \left (s -\sqrt {2}\right )}\right ) &= \frac {3 \,{\mathrm e}^{\sqrt {2}\, t}}{8}\\ \mathcal {L}^{-1}\left (\frac {3}{8 \left (s +\sqrt {2}\right )}\right ) &= \frac {3 \,{\mathrm e}^{-\sqrt {2}\, t}}{8}\\ \mathcal {L}^{-1}\left (\frac {1}{8 s -8 i \sqrt {2}}\right ) &= \frac {{\mathrm e}^{i \sqrt {2}\, t}}{8}\\ \mathcal {L}^{-1}\left (\frac {1}{8 s +8 i \sqrt {2}}\right ) &= \frac {{\mathrm e}^{-i \sqrt {2}\, t}}{8} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {\cos \left (\sqrt {2}\, t \right )}{4}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{4} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\cos \left (\sqrt {2}\, t \right )}{4}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {\cos \left (\sqrt {2}\, t \right )}{4}+\frac {3 \cosh \left (\sqrt {2}\, t \right )}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-4 y=0, y \left (0\right )=1, 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 \}}}}=1, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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 )=4 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 )=4 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 \\ 4 & 0 & 0 & 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 \\ 4 & 0 & 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 [\sqrt {2}, \left [\begin {array}{c} \frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ \frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I} \sqrt {2}, \left [\begin {array}{c} -\frac {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I} \sqrt {2}, \left [\begin {array}{c} \frac {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ -\frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ]\right ], \left [-\sqrt {2}, \left [\begin {array}{c} -\frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ -\frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\sqrt {2}, \left [\begin {array}{c} \frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ \frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{\sqrt {2}\, t}\cdot \left [\begin {array}{c} \frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ \frac {\sqrt {2}}{2} \\ 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 {\mathrm {I}}{4} \sqrt {2} \\ -\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 {\mathrm {I}}{4} \sqrt {2} \\ -\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 {\mathrm {I}}{4} \sqrt {2} \\ -\frac {1}{2} \\ \frac {\mathrm {I}}{2} \sqrt {2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{4} \left (\cos \left (\sqrt {2}\, t \right )-\mathrm {I} \sin \left (\sqrt {2}\, t \right )\right ) \sqrt {2} \\ -\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 {\sin \left (\sqrt {2}\, t \right ) \sqrt {2}}{4} \\ -\frac {\cos \left (\sqrt {2}\, t \right )}{2} \\ \frac {\sin \left (\sqrt {2}\, t \right ) \sqrt {2}}{2} \\ \cos \left (\sqrt {2}\, t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{4} \\ \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 {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\sqrt {2}, \left [\begin {array}{c} -\frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ -\frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{-\sqrt {2}\, t}\cdot \left [\begin {array}{c} -\frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ -\frac {\sqrt {2}}{2} \\ 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}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{\sqrt {2}\, t}\cdot \left [\begin {array}{c} \frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ \frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]+c_{4} {\mathrm e}^{-\sqrt {2}\, t}\cdot \left [\begin {array}{c} -\frac {\sqrt {2}}{4} \\ \frac {1}{2} \\ -\frac {\sqrt {2}}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{2} \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{4}-\frac {c_{3} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{4} \\ -\frac {c_{2} \cos \left (\sqrt {2}\, t \right )}{2}+\frac {c_{3} \sin \left (\sqrt {2}\, t \right )}{2} \\ \frac {c_{2} \sqrt {2}\, \sin \left (\sqrt {2}\, t \right )}{2}+\frac {c_{3} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}}{2} \\ c_{2} \cos \left (\sqrt {2}\, t \right )-c_{3} \sin \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 {\sqrt {2}\, \left (c_{3} \cos \left (\sqrt {2}\, t \right )+c_{2} \sin \left (\sqrt {2}\, t \right )-c_{1} {\mathrm e}^{\sqrt {2}\, t}+c_{4} {\mathrm e}^{-\sqrt {2}\, t}\right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {\sqrt {2}\, \left (c_{3} -c_{1} +c_{4} \right )}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2}\, \left (-c_{3} \sin \left (\sqrt {2}\, t \right ) \sqrt {2}+c_{2} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}-c_{1} {\mathrm e}^{\sqrt {2}\, t} \sqrt {2}-c_{4} {\mathrm e}^{-\sqrt {2}\, t} \sqrt {2}\right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {\sqrt {2}\, \left (\sqrt {2}\, c_{2} -\sqrt {2}\, c_{1} -c_{4} \sqrt {2}\right )}{4} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\sqrt {2}\, \left (-2 c_{3} \cos \left (\sqrt {2}\, t \right )-2 c_{2} \sin \left (\sqrt {2}\, t \right )-2 c_{1} {\mathrm e}^{\sqrt {2}\, t}+2 c_{4} {\mathrm e}^{-\sqrt {2}\, t}\right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-\frac {\sqrt {2}\, \left (-2 c_{3} -2 c_{1} +2 c_{4} \right )}{4} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {\sqrt {2}\, \left (2 c_{3} \sin \left (\sqrt {2}\, t \right ) \sqrt {2}-2 c_{2} \cos \left (\sqrt {2}\, t \right ) \sqrt {2}-2 c_{1} {\mathrm e}^{\sqrt {2}\, t} \sqrt {2}-2 c_{4} {\mathrm e}^{-\sqrt {2}\, t} \sqrt {2}\right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {\sqrt {2}\, \left (-2 \sqrt {2}\, c_{2} -2 \sqrt {2}\, c_{1} -2 c_{4} \sqrt {2}\right )}{4} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {3 \sqrt {2}}{4}, c_{2} =0, c_{3} =-\frac {\sqrt {2}}{2}, c_{4} =-\frac {3 \sqrt {2}}{4}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\cos \left (\sqrt {2}\, t \right )}{4}+\frac {3 \,{\mathrm e}^{\sqrt {2}\, t}}{8}+\frac {3 \,{\mathrm e}^{-\sqrt {2}\, t}}{8} \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: 0.516 (sec). Leaf size: 21
dsolve([diff(y(t),t$4)-4*y(t)=0,y(0) = 1, D(y)(0) = 0, (D@@2)(y)(0) = 1, (D@@3)(y)(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = \frac {\cos \left (t \sqrt {2}\right )}{4}+\frac {3 \cosh \left (t \sqrt {2}\right )}{4} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 43
DSolve[{y''''[t]-4*y[t]==0,{y[0]==1,y'[0]==0,y''[0]==1,y'''[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{8} \left (3 e^{-\sqrt {2} t}+3 e^{\sqrt {2} t}+2 \cos \left (\sqrt {2} t\right )\right ) \]