Internal problem ID [812]
Internal file name [OUTPUT/812_Sunday_June_05_2022_01_50_23_AM_98049129/index.tex]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 4.1, Higher order linear differential equations. General theory. page
173
Problem number: 1.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+3 y=t} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+3 y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+3 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\\ \lambda _3 &= \frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\\ \lambda _4 &= \frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2} \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-t}+{\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t} c_{2} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{3} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-t} \\ y_2 &= {\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t} \\ y_3 &= {\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \\ y_4 &= {\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+3 y = t \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ t \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, t\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t}, {\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}, {\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}, {\mathrm e}^{-t}\right \} \] 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. \[ y_p = A_{2} t +A_{1} \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ 3 A_{2} t +3 A_{1} = t \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{3}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {t}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-t}+{\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t} c_{2} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{3} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{4}\right ) + \left (\frac {t}{3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-t}+{\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t} c_{2} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{3} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{4} +\frac {t}{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-t}+{\mathrm e}^{\left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1\right ) t} c_{2} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1+\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{3} +{\mathrm e}^{\left (\frac {\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}{2}+\frac {1}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}-1-\frac {i \sqrt {3}\, \left (-\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+\frac {2}{\left (4+2 \sqrt {2}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} c_{4} +\frac {t}{3} \] 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] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 182
dsolve(diff(y(t),t$4)+4*diff(y(t),t$3)+3*y(t)=t,y(t), singsol=all)
\[ y \left (t \right ) = \frac {t}{3}+{\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{\frac {t \left (\left (\sqrt {2}-2\right ) \left (4+2 \sqrt {2}\right )^{\frac {2}{3}}-2 \left (4+2 \sqrt {2}\right )^{\frac {1}{3}}-2\right )}{2}}+c_{3} {\mathrm e}^{-\frac {t \left (\left (\sqrt {2}-2\right ) \left (4+2 \sqrt {2}\right )^{\frac {2}{3}}-2 \left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+4\right )}{4}} \cos \left (\frac {t \left (4+2 \sqrt {2}\right )^{\frac {1}{3}} \left (2+\left (\sqrt {2}-2\right ) \left (4+2 \sqrt {2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{4}\right )+c_{4} {\mathrm e}^{-\frac {t \left (\left (\sqrt {2}-2\right ) \left (4+2 \sqrt {2}\right )^{\frac {2}{3}}-2 \left (4+2 \sqrt {2}\right )^{\frac {1}{3}}+4\right )}{4}} \sin \left (\frac {t \left (4+2 \sqrt {2}\right )^{\frac {1}{3}} \left (2+\left (\sqrt {2}-2\right ) \left (4+2 \sqrt {2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{4}\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 100
DSolve[y''''[t]+4*y'''[t]+3*y[t]==t,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to c_2 \exp \left (t \text {Root}\left [\text {$\#$1}^3+3 \text {$\#$1}^2-3 \text {$\#$1}+3\&,2\right ]\right )+c_3 \exp \left (t \text {Root}\left [\text {$\#$1}^3+3 \text {$\#$1}^2-3 \text {$\#$1}+3\&,3\right ]\right )+c_1 \exp \left (t \text {Root}\left [\text {$\#$1}^3+3 \text {$\#$1}^2-3 \text {$\#$1}+3\&,1\right ]\right )+\frac {t}{3}+c_4 e^{-t} \]