Internal problem ID [14678]
Internal file name [OUTPUT/14358_Wednesday_April_03_2024_02_17_37_PM_98265897/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: 4.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+9 y^{\prime \prime }=1} \] 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 \[ y^{\prime \prime \prime \prime }+9 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+9 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 3 i\\ \lambda _4 &= -3 i \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{2} t +c_{1} +{\mathrm e}^{-3 i t} c_{3} +{\mathrm e}^{3 i t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= t \\ y_3 &= {\mathrm e}^{-3 i t} \\ y_4 &= {\mathrm e}^{3 i t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+9 y^{\prime \prime } = 1 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, t, {\mathrm e}^{-3 i t}, {\mathrm e}^{3 i t}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t\}] \] Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t^{2}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1} t^{2} \] 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 \[ 18 A_{1} = 1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{18}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {t^{2}}{18} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} t +c_{1} +{\mathrm e}^{-3 i t} c_{3} +{\mathrm e}^{3 i t} c_{4}\right ) + \left (\frac {t^{2}}{18}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} t +c_{1} +{\mathrm e}^{-3 i t} c_{3} +{\mathrm e}^{3 i t} c_{4} +\frac {t^{2}}{18} \\ \end{align*}
Verification of solutions
\[ y = c_{2} t +c_{1} +{\mathrm e}^{-3 i t} c_{3} +{\mathrm e}^{3 i t} c_{4} +\frac {t^{2}}{18} \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -9*_b(_a)+1, _b(_a)` *** Sublevel 2 *** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful <- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 28
dsolve(diff(y(t),t$4)+9*diff(y(t),t$2)=1,y(t), singsol=all)
\[ y \left (t \right ) = \frac {t^{2}}{18}-\frac {c_{1} \cos \left (3 t \right )}{9}-\frac {c_{2} \sin \left (3 t \right )}{9}+c_{3} t +c_{4} \]
✓ Solution by Mathematica
Time used: 0.16 (sec). Leaf size: 39
DSolve[y''''[t]+9*y''[t]==1,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {t^2}{18}+c_4 t-\frac {1}{9} c_1 \cos (3 t)-\frac {1}{9} c_2 \sin (3 t)+c_3 \]