Internal problem ID [15266]
Internal file name [OUTPUT/15267_Wednesday_May_08_2024_03_54_32_PM_27709327/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. Trial and error method. Exercises page 132
Problem number: 497.
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, _missing_x]]
\[ \boxed {y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }=4} \] 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 \[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}-\lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{4} \] 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 &= {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime } = 4 \] 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 \[ \{1, x, x^{2}, {\mathrm e}^{x}\} \] 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 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. \[ y_p = A_{1} x^{3} \] 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 \[ -6 A_{1} = 4 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {2}{3}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {2 x^{3}}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{4}\right ) + \left (-\frac {2 x^{3}}{3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{4} -\frac {2 x^{3}}{3} \\ \end{align*}
Verification of solutions
\[ y = x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{x} c_{4} -\frac {2 x^{3}}{3} \] 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) = _b(_a)+4, _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: 24
dsolve(diff(y(x),x$4)-diff(y(x),x$3)=4,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{x}+\frac {c_{2} x^{2}}{2}-\frac {2 x^{3}}{3}+c_{3} x +c_{4} \]
✓ Solution by Mathematica
Time used: 0.027 (sec). Leaf size: 31
DSolve[y''''[x]-y'''[x]==4,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {2 x^3}{3}+c_4 x^2+c_3 x+c_1 e^x+c_2 \]