Internal problem ID [12224]
Internal file name [OUTPUT/10877_Thursday_September_28_2023_01_05_45_AM_32247685/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Differential Equations. Problems page
221
Problem number: Problem 1(d).
ODE order: 5.
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_y]]
\[ \boxed {y^{\left (5\right )}-y^{\prime \prime \prime \prime }+y^{\prime }=2 x^{2}+3} \] 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 \[ y^{\left (5\right )}-y^{\prime \prime \prime \prime }+y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{5}-\lambda ^{4}+\lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right )\\ \lambda _3 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right )\\ \lambda _4 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right )\\ \lambda _5 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right ) x} c_{4} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=3\right ) x} \\ y_3 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=2\right ) x} \\ y_4 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=1\right ) x} \\ y_5 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=4\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (5\right )}-y^{\prime \prime \prime \prime }+y^{\prime } = 2 x^{2}+3 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x^{2}+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}, x^{3}\}] \] 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. \[ y_p = A_{3} x^{3}+A_{2} x^{2}+A_{1} x \] 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 \[ 3 x^{2} A_{3}+2 x A_{2}+A_{1} = 2 x^{2}+3 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 3, A_{2} = 0, A_{3} = {\frac {2}{3}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {2}{3} x^{3}+3 x \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=3\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=1\right ) x} c_{4} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=4\right ) x} c_{5}\right ) + \left (\frac {2}{3} x^{3}+3 x\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=3\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=1\right ) x} c_{4} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} &=4\right ) x} c_{5} +\frac {2 x^{3}}{3}+3 x \\ \end{align*}
Verification of solutions
\[ y = c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right ) x} c_{4} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) x} c_{5} +\frac {2 x^{3}}{3}+3 x \] 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: 5; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(diff(diff(_b(_a), _a), _a), _a), _a) = 2*_a^2-_b(_a)+diff(diff(diff(_b(_a), _a), _a), _a 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 <- differential order: 5; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 372
dsolve(diff(y(x),x$5)-diff(y(x),x$4) +diff(y(x),x)=2*x^2+3,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right )^{2} \left (\frac {3 c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right ) x} \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right )}{2}+\left (\frac {3 c_{2} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) x} \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right )}{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right ) \left (\frac {3 c_{3} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right ) x} \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right )}{2}+\left (\frac {3 c_{4} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) x}}{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right ) \left (x^{3}+\frac {9}{2} x +\frac {3}{2} c_{5} \right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right )\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right )\right ) \left (\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right )-1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right )^{2} \left (\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =4\right )-1\right ) \left (\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =2\right )-1\right ) \left (\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+1, \operatorname {index} =1\right )-1\right )}{3} \]
✓ Solution by Mathematica
Time used: 0.093 (sec). Leaf size: 182
DSolve[y'''''[x]-y''''[x] +y'[x]==2*x^2+3,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_2 \exp \left (x \text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,2\right ]\right )}{\text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,2\right ]}+\frac {c_1 \exp \left (x \text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,1\right ]\right )}{\text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,1\right ]}+\frac {c_4 \exp \left (x \text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,4\right ]\right )}{\text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,4\right ]}+\frac {c_3 \exp \left (x \text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,3\right ]\right )}{\text {Root}\left [\text {$\#$1}^4-\text {$\#$1}^3+1\&,3\right ]}+\frac {2 x^3}{3}+3 x+c_5 \]