Internal problem ID [14660]
Internal file name [OUTPUT/14340_Wednesday_April_03_2024_02_17_23_PM_46263500/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: 42.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+16 y^{\prime \prime \prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 1] \end {align*}
The characteristic equation is \[ \lambda ^{4}+16 \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -16\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-16 t} c_{4} \] 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 &= {\mathrm e}^{-16 t} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = t^{2} c_{3} +c_{2} t +c_{1} +{\mathrm e}^{-16 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 \(y = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{3} t +c_{2} -16 \,{\mathrm e}^{-16 t} c_{4} \end {align*}
substituting \(y^{\prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = c_{2} -16 c_{4}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 2 c_{3} +256 \,{\mathrm e}^{-16 t} c_{4} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = 2 c_{3} +256 c_{4}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -4096 \,{\mathrm e}^{-16 t} c_{4} \end {align*}
substituting \(y^{\prime \prime \prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = -4096 c_{4}\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}&={\frac {1}{4096}}\\ c_{2}&={\frac {255}{256}}\\ c_{3}&={\frac {1}{32}}\\ c_{4}&=-{\frac {1}{4096}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {1}{4096}+\frac {t^{2}}{32}+\frac {255 t}{256}-\frac {{\mathrm e}^{-16 t}}{4096} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{4096}+\frac {t^{2}}{32}+\frac {255 t}{256}-\frac {{\mathrm e}^{-16 t}}{4096} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{4096}+\frac {t^{2}}{32}+\frac {255 t}{256}-\frac {{\mathrm e}^{-16 t}}{4096} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }+16 y^{\prime \prime \prime }=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 \}}}}=0, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\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_{4}\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_{4}\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 \\ 0 & 0 & 0 & -16 \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 \\ 0 & 0 & 0 & -16 \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 [-16, \left [\begin {array}{c} -\frac {1}{4096} \\ \frac {1}{256} \\ -\frac {1}{16} \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-16, \left [\begin {array}{c} -\frac {1}{4096} \\ \frac {1}{256} \\ -\frac {1}{16} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-16 t}\cdot \left [\begin {array}{c} -\frac {1}{4096} \\ \frac {1}{256} \\ -\frac {1}{16} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 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 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 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 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 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 \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} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-16 t}\cdot \left [\begin {array}{c} -\frac {1}{4096} \\ \frac {1}{256} \\ -\frac {1}{16} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 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}^{-16 t}}{4096}+c_{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\frac {c_{1}}{4096}+c_{2} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1} {\mathrm e}^{-16 t}}{256} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=\frac {c_{1}}{256} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {c_{1} {\mathrm e}^{-16 t}}{16} \\ \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}}{16} \\ \bullet & {} & \textrm {Calculate the 3rd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=c_{1} {\mathrm e}^{-16 t} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=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: 0.016 (sec). Leaf size: 20
dsolve([diff(y(t),t$4)+16*diff(y(t),t$3)=0,y(0) = 0, D(y)(0) = 1, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 1],y(t), singsol=all)
\[ y \left (t \right ) = \frac {1}{4096}+\frac {255 t}{256}+\frac {t^{2}}{32}-\frac {{\mathrm e}^{-16 t}}{4096} \]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 26
DSolve[{y''''[t]+16*y'''[t]==0,{y[0]==0,y'[0]==1,y''[0]==0,y'''[0]==1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {128 t^2+4080 t-e^{-16 t}+1}{4096} \]