Internal problem ID [9905]
Internal file name [OUTPUT/8852_Monday_June_06_2022_05_40_33_AM_84889281/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 5, linear fifth and higher order
Problem number: 1583.
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_x]]
\[ \boxed {y^{\left (5\right )}+a y^{\prime \prime \prime \prime }=f} \] 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 \[ y^{\left (5\right )}+a y^{\prime \prime \prime \prime } = 0 \] The characteristic equation is \[ a \,\lambda ^{4}+\lambda ^{5} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -a\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0\\ \lambda _5 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-a x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= x^{2} \\ y_4 &= x^{3} \\ y_5 &= {\mathrm e}^{-a x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (5\right )}+a y^{\prime \prime \prime \prime } = f \] 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 \[ \{1, x, x^{2}, x^{3}, {\mathrm e}^{-a x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2}\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3}\}] \] Since \(x^{3}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{4}\}] \] 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. \[ y_p = A_{1} x^{4} \] 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 \[ 24 a A_{1} = f \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = \frac {f}{24 a}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {f \,x^{4}}{24 a} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-a x} c_{5}\right ) + \left (\frac {f \,x^{4}}{24 a}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-a x} c_{5} +\frac {f \,x^{4}}{24 a} \\ \end{align*}
Verification of solutions
\[ y = c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-a x} c_{5} +\frac {f \,x^{4}}{24 a} \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -a*_b(_a)+f, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 50
dsolve(diff(y(x),x$5)+a*diff(y(x),x$4)-f=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {6 \,{\mathrm e}^{-a x} c_{1} +a^{3} \left (\left (c_{2} x^{3}+3 x^{2} c_{3} +6 c_{4} x +6 c_{5} \right ) a +\frac {f \,x^{4}}{4}\right )}{6 a^{4}} \]
✓ Solution by Mathematica
Time used: 0.222 (sec). Leaf size: 45
DSolve[y'''''[x]+a*y''''[x]-f==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1 e^{-a x}}{a^4}+\frac {f x^4}{24 a}+x (x (c_5 x+c_4)+c_3)+c_2 \]