Internal problem ID [14857]
Internal file name [OUTPUT/14537_Monday_April_08_2024_06_26_44_AM_8663531/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 37.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+14 y^{\prime \prime }+20 y^{\prime }+25 y=t^{2}} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+14 y^{\prime \prime }+20 y^{\prime }+25 y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+14 \lambda ^{2}+20 \lambda +25 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1-2 i\\ \lambda _2 &= -1+2 i\\ \lambda _3 &= -1-2 i\\ \lambda _4 &= -1+2 i \end {align*}
Therefore the homogeneous solution is \[ y_h(t)={\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +t \,{\mathrm e}^{\left (-1+2 i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +t \,{\mathrm e}^{\left (-1-2 i\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\left (-1+2 i\right ) t} \\ y_2 &= t \,{\mathrm e}^{\left (-1+2 i\right ) t} \\ y_3 &= {\mathrm e}^{\left (-1-2 i\right ) t} \\ y_4 &= t \,{\mathrm e}^{\left (-1-2 i\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+14 y^{\prime \prime }+20 y^{\prime }+25 y = t^{2} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ t^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, t, t^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{t \,{\mathrm e}^{\left (-1-2 i\right ) t}, t \,{\mathrm e}^{\left (-1+2 i\right ) t}, {\mathrm e}^{\left (-1-2 i\right ) t}, {\mathrm e}^{\left (-1+2 i\right ) t}\} \] 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. \[ y_p = A_{3} t^{2}+A_{2} t +A_{1} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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 \[ 25 A_{3} t^{2}+25 A_{2} t +40 t A_{3}+25 A_{1}+20 A_{2}+28 A_{3} = t^{2} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {4}{625}}, A_{2} = -{\frac {8}{125}}, A_{3} = {\frac {1}{25}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {1}{25} t^{2}-\frac {8}{125} t +\frac {4}{625} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\left (-1+2 i\right ) t} c_{1} +t \,{\mathrm e}^{\left (-1+2 i\right ) t} c_{2} +{\mathrm e}^{\left (-1-2 i\right ) t} c_{3} +t \,{\mathrm e}^{\left (-1-2 i\right ) t} c_{4}\right ) + \left (\frac {1}{25} t^{2}-\frac {8}{125} t +\frac {4}{625}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (-1-2 i\right ) t}+{\mathrm e}^{\left (-1+2 i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {t^{2}}{25}-\frac {8 t}{125}+\frac {4}{625} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (-1-2 i\right ) t}+{\mathrm e}^{\left (-1+2 i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {t^{2}}{25}-\frac {8 t}{125}+\frac {4}{625} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (-1-2 i\right ) t}+{\mathrm e}^{\left (-1+2 i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {t^{2}}{25}-\frac {8 t}{125}+\frac {4}{625} \] 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: 0.015 (sec). Leaf size: 40
dsolve(diff(y(t),t$4)+4*diff(y(t),t$3)+14*diff(y(t),t$2)+20*diff(y(t),t)+25*y(t)=t^2,y(t), singsol=all)
\[ y \left (t \right ) = \frac {4}{625}+\left (\left (c_{3} t +c_{1} \right ) \cos \left (2 t \right )+\sin \left (2 t \right ) \left (t c_{4} +c_{2} \right )\right ) {\mathrm e}^{-t}+\frac {t^{2}}{25}-\frac {8 t}{125} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 54
DSolve[y''''[t]+4*y'''[t]+14*y''[t]+20*y'[t]+25*y[t]==t^2,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {1}{625} \left (25 t^2-40 t+4\right )+e^{-t} (c_4 t+c_3) \cos (2 t)+e^{-t} (c_2 t+c_1) \sin (2 t) \]