Internal problem ID [15357]
Internal file name [OUTPUT/15358_Wednesday_May_08_2024_03_56_55_PM_69386188/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.3 Nonhomogeneous linear equations with
constant coefficients. Superposition principle. Exercises page 137
Problem number: 589.
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]]
\[ \boxed {y^{\left (5\right )}-y^{\prime \prime \prime }=x +2 \,{\mathrm e}^{-x}} \] 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 \[ y^{\left (5\right )}-y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{5}-\lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 1\\ \lambda _5 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +x^{2} c_{4} +{\mathrm e}^{x} c_{5} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= x \\ y_4 &= x^{2} \\ y_5 &= {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (5\right )}-y^{\prime \prime \prime } = x +2 \,{\mathrm e}^{-x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x +2 \,{\mathrm e}^{-x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x}\}, \{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, x^{2}, {\mathrm e}^{x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} x\}, \{1, x\}] \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} x\}, \{x, x^{2}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} x\}, \{x^{2}, x^{3}\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} x\}, \{x^{3}, x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-x} x +A_{2} x^{3}+A_{3} x^{4} \] 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 \[ 2 A_{1} {\mathrm e}^{-x}-6 A_{2}-24 A_{3} x = x +2 \,{\mathrm e}^{-x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 1, A_{2} = 0, A_{3} = -{\frac {1}{24}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{-x} x -\frac {x^{4}}{24} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +x^{2} c_{4} +{\mathrm e}^{x} c_{5}\right ) + \left ({\mathrm e}^{-x} x -\frac {x^{4}}{24}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +x^{2} c_{4} +{\mathrm e}^{x} c_{5} +{\mathrm e}^{-x} x -\frac {x^{4}}{24} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +x^{2} c_{4} +{\mathrm e}^{x} c_{5} +{\mathrm e}^{-x} x -\frac {x^{4}}{24} \] 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(_b(_a), _a), _a) = _b(_a)+_a+2*exp(-_a), _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: 5; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve(diff(y(x),x$5)-diff(y(x),x$3)=x+2*exp(-x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (7+2 x -2 c_{1} \right ) {\mathrm e}^{-x}}{2}-\frac {x^{4}}{24}+\frac {c_{3} x^{2}}{2}+c_{4} x +c_{2} {\mathrm e}^{x}+c_{5} \]
✓ Solution by Mathematica
Time used: 0.394 (sec). Leaf size: 46
DSolve[y'''''[x]-y'''[x]==x+2*Exp[-x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x^4}{24}+c_5 x^2+c_4 x+c_1 e^x+e^{-x} \left (x+\frac {7}{2}-c_2\right )+c_3 \]