Internal problem ID [14667]
Internal file name [OUTPUT/14347_Wednesday_April_03_2024_02_17_25_PM_53610816/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: 49.
ODE order: 5.
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]]
\[ \boxed {y^{\left (5\right )}+8 y^{\prime \prime \prime \prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 8, y^{\prime }\left (0\right ) = 4, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 48, y^{\prime \prime \prime \prime }\left (0\right ) = 0] \end {align*}
The characteristic equation is \[ \lambda ^{5}+8 \lambda ^{4} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -8\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0\\ \lambda _5 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=t^{3} c_{4} +t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-8 t} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= t\\ y_3 &= t^{2}\\ y_4 &= t^{3}\\ y_5 &= {\mathrm e}^{-8 t} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = t^{3} c_{4} +t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-8 t} c_{5} \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 \(y = 8\) and \(t = 0\) in the above gives \begin {align*} 8 = c_{1} +c_{5}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 3 t^{2} c_{4} +2 c_{3} t +c_{2} -8 \,{\mathrm e}^{-8 t} c_{5} \end {align*}
substituting \(y^{\prime } = 4\) and \(t = 0\) in the above gives \begin {align*} 4 = c_{2} -8 c_{5}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 6 t c_{4} +2 c_{3} +64 \,{\mathrm e}^{-8 t} c_{5} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = 2 c_{3} +64 c_{5}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = 6 c_{4} -512 \,{\mathrm e}^{-8 t} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime } = 48\) and \(t = 0\) in the above gives \begin {align*} 48 = 6 c_{4} -512 c_{5}\tag {4A} \end {align*}
Taking four derivatives of the solution gives \begin {align*} y^{\prime \prime \prime \prime } = 4096 \,{\mathrm e}^{-8 t} c_{5} \end {align*}
substituting \(y^{\prime \prime \prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = 4096 c_{5}\tag {5A} \end {align*}
Equations {1A,2A,3A,4A,5A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}\}\). Solving for the constants gives \begin {align*} c_{1}&=8\\ c_{2}&=4\\ c_{3}&=0\\ c_{4}&=8\\ c_{5}&=0 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = 8 t^{3}+4 t +8 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 8 t^{3}+4 t +8 \\ \end{align*}
Verification of solutions
\[ y = 8 t^{3}+4 t +8 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\left (5\right )}+8 y^{\prime \prime \prime \prime }=0, y \left (0\right )=8, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=48, y^{\prime \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} 5 \\ {} & {} & y^{\left (5\right )} \\ \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 {Define new variable}\hspace {3pt} y_{5}\left (t \right ) \\ {} & {} & y_{5}\left (t \right )=y^{\prime \prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{5}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{5}^{\prime }\left (t \right )=-8 y_{5}\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_{5}\left (t \right )=y_{4}^{\prime }\left (t \right ), y_{5}^{\prime }\left (t \right )=-8 y_{5}\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 ) \\ y_{5}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -8 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -8 \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 [-8, \left [\begin {array}{c} \frac {1}{4096} \\ -\frac {1}{512} \\ \frac {1}{64} \\ -\frac {1}{8} \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-8, \left [\begin {array}{c} \frac {1}{4096} \\ -\frac {1}{512} \\ \frac {1}{64} \\ -\frac {1}{8} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-8 t}\cdot \left [\begin {array}{c} \frac {1}{4096} \\ -\frac {1}{512} \\ \frac {1}{64} \\ -\frac {1}{8} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{5}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \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}+c_{3} {\moverset {\rightarrow }{y}}_{3}+c_{4} {\moverset {\rightarrow }{y}}_{4}+c_{5} {\moverset {\rightarrow }{y}}_{5} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-8 t}\cdot \left [\begin {array}{c} \frac {1}{4096} \\ -\frac {1}{512} \\ \frac {1}{64} \\ -\frac {1}{8} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{1} {\mathrm e}^{-8 t}}{4096}+c_{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=8 \\ {} & {} & 8=\frac {c_{1}}{4096}+c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{-8 t}}{512} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=-\frac {c_{1}}{512} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {c_{1} {\mathrm e}^{-8 t}}{64} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\frac {c_{1}}{64} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {c_{1} {\mathrm e}^{-8 t}}{8} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=48 \\ {} & {} & 48=-\frac {c_{1}}{8} \\ \bullet & {} & \textrm {Calculate the 4th derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=c_{1} {\mathrm e}^{-8 t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ \bullet & {} & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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: 0.031 (sec). Leaf size: 14
dsolve([diff(y(t),t$5)+8*diff(y(t),t$4)=0,y(0) = 8, D(y)(0) = 4, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 48, (D@@4)(y)(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = 8 t^{3}+4 t +8 \]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 15
DSolve[{y'''''[t]+8*y''''[t]==0,{y[0]==8,y'[0]==4,y''[0]==0,y'''[0]==48,y''''[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to 4 \left (2 t^3+t+2\right ) \]