Internal problem ID [14664]
Internal file name [OUTPUT/14344_Wednesday_April_03_2024_02_17_24_PM_51780437/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: 46.
ODE order: 4.
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
[[_high_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime \prime }-16 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2, y^{\prime \prime }\left (0\right ) = 4, y^{\prime \prime \prime }\left (0\right ) = -24] \end {align*}
The characteristic equation is \[ \lambda ^{4}-16 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= 2 i\\ \lambda _4 &= -2 i \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-2 t}\\ y_2 &= {\mathrm e}^{2 t}\\ y_3 &= {\mathrm e}^{-2 i t}\\ y_4 &= {\mathrm e}^{2 i 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}^{-2 t}+c_{2} {\mathrm e}^{2 t}+{\mathrm e}^{-2 i t} c_{3} +{\mathrm e}^{2 i t} c_{4} \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 \(y = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = c_{1} +c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -2 c_{1} {\mathrm e}^{-2 t}+2 c_{2} {\mathrm e}^{2 t}-2 i {\mathrm e}^{-2 i t} c_{3} +2 i {\mathrm e}^{2 i t} c_{4} \end {align*}
substituting \(y^{\prime } = 2\) and \(t = 0\) in the above gives \begin {align*} 2 = -2 c_{3} i+2 c_{4} i-2 c_{1} +2 c_{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 4 c_{1} {\mathrm e}^{-2 t}+4 c_{2} {\mathrm e}^{2 t}-4 \,{\mathrm e}^{-2 i t} c_{3} -4 \,{\mathrm e}^{2 i t} c_{4} \end {align*}
substituting \(y^{\prime \prime } = 4\) and \(t = 0\) in the above gives \begin {align*} 4 = 4 c_{1} +4 c_{2} -4 c_{3} -4 c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -8 c_{1} {\mathrm e}^{-2 t}+8 c_{2} {\mathrm e}^{2 t}+8 i {\mathrm e}^{-2 i t} c_{3} -8 i {\mathrm e}^{2 i t} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = -24\) and \(t = 0\) in the above gives \begin {align*} -24 = 8 c_{3} i-8 c_{4} i-8 c_{1} +8 c_{2}\tag {4A} \end {align*}
Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=0\\ c_{3}&=i\\ c_{4}&=-i \end {align*}
Substituting these values back in above solution results in \begin {align*} y = {\mathrm e}^{-2 t}+2 \sin \left (2 t \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 t}+2 \sin \left (2 t \right ) \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-2 t}+2 \sin \left (2 t \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-16 y=0, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-24\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 )=16 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 )=16 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 \\ 16 & 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 \\ 16 & 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 [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{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 [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\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}+c_{2} {\moverset {\rightarrow }{y}}_{2}+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}}=c_{1} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{3} \sin \left (2 t \right )}{8}-\frac {c_{4} \cos \left (2 t \right )}{8} \\ -\frac {c_{3} \cos \left (2 t \right )}{4}+\frac {c_{4} \sin \left (2 t \right )}{4} \\ \frac {c_{3} \sin \left (2 t \right )}{2}+\frac {c_{4} \cos \left (2 t \right )}{2} \\ c_{3} \cos \left (2 t \right )-c_{4} \sin \left (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}^{-2 t}}{8}+\frac {c_{2} {\mathrm e}^{2 t}}{8}-\frac {c_{4} \cos \left (2 t \right )}{8}-\frac {c_{3} \sin \left (2 t \right )}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {c_{1}}{8}+\frac {c_{2}}{8}-\frac {c_{4}}{8} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1} {\mathrm e}^{-2 t}}{4}+\frac {c_{2} {\mathrm e}^{2 t}}{4}+\frac {c_{4} \sin \left (2 t \right )}{4}-\frac {c_{3} \cos \left (2 t \right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=\frac {c_{1}}{4}+\frac {c_{2}}{4}-\frac {c_{3}}{4} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {c_{1} {\mathrm e}^{-2 t}}{2}+\frac {c_{2} {\mathrm e}^{2 t}}{2}+\frac {c_{4} \cos \left (2 t \right )}{2}+\frac {c_{3} \sin \left (2 t \right )}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=-\frac {c_{1}}{2}+\frac {c_{2}}{2}+\frac {c_{4}}{2} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{2 t}-c_{4} \sin \left (2 t \right )+c_{3} \cos \left (2 t \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-24 \\ {} & {} & -24=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-8, c_{2} =0, c_{3} =-16, c_{4} =0\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-2 t}+2 \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: 0.015 (sec). Leaf size: 15
dsolve([diff(y(t),t$4)-16*y(t)=0,y(0) = 1, D(y)(0) = 2, (D@@2)(y)(0) = 4, (D@@3)(y)(0) = -24],y(t), singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{-2 t}+2 \sin \left (2 t \right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 17
DSolve[{y''''[t]-16*y[t]==0,{y[0]==1,y'[0]==2,y''[0]==4,y'''[0]==-24}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-2 t}+2 \sin (2 t) \]