Internal problem ID [15356]
Internal file name [OUTPUT/15357_Wednesday_May_08_2024_03_56_54_PM_31214681/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: 588.
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]]
\[ \boxed {y^{\left (5\right )}-y^{\prime \prime \prime \prime }={\mathrm e}^{x} x -1} \] 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 \[ y^{\left (5\right )}-y^{\prime \prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{5}-\lambda ^{4} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0\\ \lambda _5 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=x^{3} c_{4} +x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{5} \] 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}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (5\right )}-y^{\prime \prime \prime \prime } = {\mathrm e}^{x} x -1 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} x -1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, x^{2}, x^{3}, {\mathrm e}^{x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \(x^{3}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{4}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{4}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}\}] \] 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. \[ y_p = A_{1} x^{4}+A_{2} {\mathrm e}^{x} x +A_{3} {\mathrm e}^{x} x^{2} \] 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 \[ A_{2} {\mathrm e}^{x}+2 A_{3} {\mathrm e}^{x} x +8 A_{3} {\mathrm e}^{x}-24 A_{1} = {\mathrm e}^{x} x -1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{24}}, A_{2} = -4, A_{3} = {\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{4}}{24}-4 \,{\mathrm e}^{x} x +\frac {{\mathrm e}^{x} x^{2}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x^{3} c_{4} +x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{5}\right ) + \left (\frac {x^{4}}{24}-4 \,{\mathrm e}^{x} x +\frac {{\mathrm e}^{x} x^{2}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{3} c_{4} +x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{5} +\frac {x^{4}}{24}-4 \,{\mathrm e}^{x} x +\frac {{\mathrm e}^{x} x^{2}}{2} \\ \end{align*}
Verification of solutions
\[ y = x^{3} c_{4} +x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{5} +\frac {x^{4}}{24}-4 \,{\mathrm e}^{x} x +\frac {{\mathrm e}^{x} x^{2}}{2} \] 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) = exp(_a)*_a+_b(_a)-1, _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: 41
dsolve(diff(y(x),x$5)-diff(y(x),x$4)=x*exp(x)-1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{2}+2 c_{1} -8 x +20\right ) {\mathrm e}^{x}}{2}+\frac {x^{4}}{24}+\frac {c_{2} x^{3}}{6}+\frac {c_{3} x^{2}}{2}+c_{4} x +c_{5} \]
✓ Solution by Mathematica
Time used: 0.337 (sec). Leaf size: 49
DSolve[y'''''[x]-y''''[x]==x*Exp[x]-1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^4}{24}+c_5 x^3+c_4 x^2+e^x \left (\frac {x^2}{2}-4 x+10+c_1\right )+c_3 x+c_2 \]