Internal problem ID [14694]
Internal file name [OUTPUT/14374_Wednesday_April_03_2024_02_17_49_PM_36170570/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.6, page 187
Problem number: 20.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime }-4 y^{\prime \prime }-11 y^{\prime }+30 y={\mathrm e}^{4 t}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime }-4 y^{\prime \prime }-11 y^{\prime }+30 y = 0 \] The characteristic equation is \[ \lambda ^{3}-4 \lambda ^{2}-11 \lambda +30 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 5\\ \lambda _3 &= -3 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{5 t} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-3 t} \\ y_2 &= {\mathrm e}^{2 t} \\ y_3 &= {\mathrm e}^{5 t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-4 y^{\prime \prime }-11 y^{\prime }+30 y = {\mathrm e}^{4 t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{4 t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{4 t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{-3 t}, {\mathrm e}^{2 t}, {\mathrm e}^{5 t}\} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} {\mathrm e}^{4 t} \] The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ -14 A_{1} {\mathrm e}^{4 t} = {\mathrm e}^{4 t} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{14}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{4 t}}{14} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{5 t}\right ) + \left (-\frac {{\mathrm e}^{4 t}}{14}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{5 t}-\frac {{\mathrm e}^{4 t}}{14} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-3 t}+c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{5 t}-\frac {{\mathrm e}^{4 t}}{14} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-4 y^{\prime \prime }-11 y^{\prime }+30 y={\mathrm e}^{4 t} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (t \right )={\mathrm e}^{4 t}+4 y_{3}\left (t \right )+11 y_{2}\left (t \right )-30 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_{3}^{\prime }\left (t \right )={\mathrm e}^{4 t}+4 y_{3}\left (t \right )+11 y_{2}\left (t \right )-30 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 ) \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 \\ -30 & 11 & 4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{4 t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{4 t} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -30 & 11 & 4 \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 )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-3, \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [5, \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-3, \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \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}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [5, \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{5 t}\cdot \left [\begin {array}{c} \frac {1}{25} \\ \frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-3 t}}{9} & \frac {{\mathrm e}^{2 t}}{4} & \frac {{\mathrm e}^{5 t}}{25} \\ -\frac {{\mathrm e}^{-3 t}}{3} & \frac {{\mathrm e}^{2 t}}{2} & \frac {{\mathrm e}^{5 t}}{5} \\ {\mathrm e}^{-3 t} & {\mathrm e}^{2 t} & {\mathrm e}^{5 t} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-3 t}}{9} & \frac {{\mathrm e}^{2 t}}{4} & \frac {{\mathrm e}^{5 t}}{25} \\ -\frac {{\mathrm e}^{-3 t}}{3} & \frac {{\mathrm e}^{2 t}}{2} & \frac {{\mathrm e}^{5 t}}{5} \\ {\mathrm e}^{-3 t} & {\mathrm e}^{2 t} & {\mathrm e}^{5 t} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{9} & \frac {1}{4} & \frac {1}{25} \\ -\frac {1}{3} & \frac {1}{2} & \frac {1}{5} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{ccc} -\frac {\left ({\mathrm e}^{8 t}-4 \,{\mathrm e}^{5 t}-1\right ) {\mathrm e}^{-3 t}}{4} & \frac {\left (5 \,{\mathrm e}^{8 t}+16 \,{\mathrm e}^{5 t}-21\right ) {\mathrm e}^{-3 t}}{120} & \frac {\left (5 \,{\mathrm e}^{8 t}-8 \,{\mathrm e}^{5 t}+3\right ) {\mathrm e}^{-3 t}}{120} \\ -\frac {\left (5 \,{\mathrm e}^{8 t}-8 \,{\mathrm e}^{5 t}+3\right ) {\mathrm e}^{-3 t}}{4} & \frac {\left (25 \,{\mathrm e}^{8 t}+32 \,{\mathrm e}^{5 t}+63\right ) {\mathrm e}^{-3 t}}{120} & \frac {\left (25 \,{\mathrm e}^{8 t}-16 \,{\mathrm e}^{5 t}-9\right ) {\mathrm e}^{-3 t}}{120} \\ -\frac {\left (25 \,{\mathrm e}^{8 t}-16 \,{\mathrm e}^{5 t}-9\right ) {\mathrm e}^{-3 t}}{4} & \frac {\left (125 \,{\mathrm e}^{8 t}+64 \,{\mathrm e}^{5 t}-189\right ) {\mathrm e}^{-3 t}}{120} & \frac {\left (125 \,{\mathrm e}^{8 t}-32 \,{\mathrm e}^{5 t}+27\right ) {\mathrm e}^{-3 t}}{120} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\left [\begin {array}{c} \frac {\left (35 \,{\mathrm e}^{8 t}-60 \,{\mathrm e}^{7 t}+28 \,{\mathrm e}^{5 t}-3\right ) {\mathrm e}^{-3 t}}{840} \\ \frac {\left (175 \,{\mathrm e}^{8 t}-240 \,{\mathrm e}^{7 t}+56 \,{\mathrm e}^{5 t}+9\right ) {\mathrm e}^{-3 t}}{840} \\ \frac {\left (875 \,{\mathrm e}^{8 t}-960 \,{\mathrm e}^{7 t}+112 \,{\mathrm e}^{5 t}-27\right ) {\mathrm e}^{-3 t}}{840} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} \frac {\left (35 \,{\mathrm e}^{8 t}-60 \,{\mathrm e}^{7 t}+28 \,{\mathrm e}^{5 t}-3\right ) {\mathrm e}^{-3 t}}{840} \\ \frac {\left (175 \,{\mathrm e}^{8 t}-240 \,{\mathrm e}^{7 t}+56 \,{\mathrm e}^{5 t}+9\right ) {\mathrm e}^{-3 t}}{840} \\ \frac {\left (875 \,{\mathrm e}^{8 t}-960 \,{\mathrm e}^{7 t}+112 \,{\mathrm e}^{5 t}-27\right ) {\mathrm e}^{-3 t}}{840} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (504 c_{3} {\mathrm e}^{8 t}+525 \,{\mathrm e}^{8 t}-900 \,{\mathrm e}^{7 t}+3150 c_{2} {\mathrm e}^{5 t}+420 \,{\mathrm e}^{5 t}+1400 c_{1} -45\right ) {\mathrm e}^{-3 t}}{12600} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 32
dsolve(diff(y(t),t$3)-4*diff(y(t),t$2)-11*diff(y(t),t)+30*y(t)=exp(4*t),y(t), singsol=all)
\[ y \left (t \right ) = -\frac {\left (-14 c_{3} {\mathrm e}^{8 t}+{\mathrm e}^{7 t}-14 c_{2} {\mathrm e}^{5 t}-14 c_{1} \right ) {\mathrm e}^{-3 t}}{14} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 39
DSolve[y'''[t]-4*y''[t]-11*y'[t]+30*y[t]==Exp[4*t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to -\frac {e^{4 t}}{14}+c_1 e^{-3 t}+c_2 e^{2 t}+c_3 e^{5 t} \]