Internal problem ID [14662]
Internal file name [OUTPUT/14342_Wednesday_April_03_2024_02_17_24_PM_50415031/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: 44.
ODE order: 3.
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
[[_3rd_order, _missing_x]]
\[ \boxed {24 y^{\prime \prime \prime }-26 y^{\prime \prime }+9 y^{\prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1, y^{\prime \prime }\left (0\right ) = 0] \end {align*}
The characteristic equation is \[ 24 \lambda ^{3}-26 \lambda ^{2}+9 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= {\frac {1}{3}}\\ \lambda _2 &= {\frac {1}{2}}\\ \lambda _3 &= {\frac {1}{4}} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{\frac {t}{3}}+{\mathrm e}^{\frac {t}{4}} c_{2} +{\mathrm e}^{\frac {t}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{\frac {t}{3}}\\ y_2 &= {\mathrm e}^{\frac {t}{4}}\\ y_3 &= {\mathrm e}^{\frac {t}{2}} \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}^{\frac {t}{3}}+{\mathrm e}^{\frac {t}{4}} c_{2} +{\mathrm e}^{\frac {t}{2}} c_{3} \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_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {c_{1} {\mathrm e}^{\frac {t}{3}}}{3}+\frac {{\mathrm e}^{\frac {t}{4}} c_{2}}{4}+\frac {{\mathrm e}^{\frac {t}{2}} c_{3}}{2} \end {align*}
substituting \(y^{\prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = \frac {c_{1}}{3}+\frac {c_{2}}{4}+\frac {c_{3}}{2}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \frac {c_{1} {\mathrm e}^{\frac {t}{3}}}{9}+\frac {{\mathrm e}^{\frac {t}{4}} c_{2}}{16}+\frac {{\mathrm e}^{\frac {t}{2}} c_{3}}{4} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = \frac {c_{1}}{9}+\frac {c_{2}}{16}+\frac {c_{3}}{4}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=54\\ c_{2}&=-40\\ c_{3}&=-14 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = 54 \,{\mathrm e}^{\frac {t}{3}}-40 \,{\mathrm e}^{\frac {t}{4}}-14 \,{\mathrm e}^{\frac {t}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 54 \,{\mathrm e}^{\frac {t}{3}}-40 \,{\mathrm e}^{\frac {t}{4}}-14 \,{\mathrm e}^{\frac {t}{2}} \\ \end{align*}
Verification of solutions
\[ y = 54 \,{\mathrm e}^{\frac {t}{3}}-40 \,{\mathrm e}^{\frac {t}{4}}-14 \,{\mathrm e}^{\frac {t}{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [24 y^{\prime \prime \prime }-26 y^{\prime \prime }+9 y^{\prime }-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 \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\frac {13 y^{\prime \prime }}{12}-\frac {3 y^{\prime }}{8}+\frac {y}{24} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-\frac {13 y^{\prime \prime }}{12}+\frac {3 y^{\prime }}{8}-\frac {y}{24}=0 \\ \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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )=\frac {13 y_{3}\left (t \right )}{12}-\frac {3 y_{2}\left (t \right )}{8}+\frac {y_{1}\left (t \right )}{24} \\ & {} & \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_{3}^{\prime }\left (t \right )=\frac {13 y_{3}\left (t \right )}{12}-\frac {3 y_{2}\left (t \right )}{8}+\frac {y_{1}\left (t \right )}{24}\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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{24} & -\frac {3}{8} & \frac {13}{12} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{24} & -\frac {3}{8} & \frac {13}{12} \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 [\frac {1}{4}, \left [\begin {array}{c} 16 \\ 4 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{3}, \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{4}, \left [\begin {array}{c} 16 \\ 4 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{\frac {t}{4}}\cdot \left [\begin {array}{c} 16 \\ 4 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{3}, \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\frac {t}{3}}\cdot \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\frac {t}{2}}\cdot \left [\begin {array}{c} 4 \\ 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}+c_{3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{\frac {t}{4}}\cdot \left [\begin {array}{c} 16 \\ 4 \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{\frac {t}{3}}\cdot \left [\begin {array}{c} 9 \\ 3 \\ 1 \end {array}\right ]+{\mathrm e}^{\frac {t}{2}} c_{3} \cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=16 c_{1} {\mathrm e}^{\frac {t}{4}}+9 c_{2} {\mathrm e}^{\frac {t}{3}}+4 \,{\mathrm e}^{\frac {t}{2}} c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=16 c_{1} +9 c_{2} +4 c_{3} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=4 c_{1} {\mathrm e}^{\frac {t}{4}}+3 c_{2} {\mathrm e}^{\frac {t}{3}}+2 \,{\mathrm e}^{\frac {t}{2}} c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=4 c_{1} +3 c_{2} +2 c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{1} {\mathrm e}^{\frac {t}{4}}+c_{2} {\mathrm e}^{\frac {t}{3}}+{\mathrm e}^{\frac {t}{2}} c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-\frac {5}{2}, c_{2} =6, c_{3} =-\frac {7}{2}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=54 \,{\mathrm e}^{\frac {t}{3}}-40 \,{\mathrm e}^{\frac {t}{4}}-14 \,{\mathrm e}^{\frac {t}{2}} \end {array} \]
Maple trace
`Methods for third 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: 23
dsolve([24*diff(y(t),t$3)-26*diff(y(t),t$2)+9*diff(y(t),t)-y(t)=0,y(0) = 0, D(y)(0) = 1, (D@@2)(y)(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = 54 \,{\mathrm e}^{\frac {t}{3}}-14 \,{\mathrm e}^{\frac {t}{2}}-40 \,{\mathrm e}^{\frac {t}{4}} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 33
DSolve[{24*y'''[t]-26*y''[t]+9*y'[t]-y[t]==0,{y[0]==0,y'[0]==1,y''[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to -40 e^{t/4}+54 e^{t/3}-14 e^{t/2} \]