Internal problem ID [15267]
Internal file name [OUTPUT/15268_Wednesday_May_08_2024_03_54_32_PM_53691888/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: 498.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+4 y^{\prime \prime }=1} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+4 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+4 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= -2\\ \lambda _4 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=x c_{2} +c_{1} +{\mathrm e}^{-2 x} c_{3} +x \,{\mathrm e}^{-2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{-2 x} \\ y_4 &= x \,{\mathrm e}^{-2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+4 y^{\prime \prime } = 1 \] 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 \[ \{1, x, x \,{\mathrm e}^{-2 x}, {\mathrm e}^{-2 x}\} \] 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 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. \[ y_p = A_{1} x^{2} \] 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 \[ 8 A_{1} = 1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{8}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{2}}{8} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x c_{2} +c_{1} +{\mathrm e}^{-2 x} c_{3} +x \,{\mathrm e}^{-2 x} c_{4}\right ) + \left (\frac {x^{2}}{8}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+x c_{2} +c_{1} +\frac {x^{2}}{8} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+x c_{2} +c_{1} +\frac {x^{2}}{8} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+x c_{2} +c_{1} +\frac {x^{2}}{8} \] 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: 4; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -4*(diff(_b(_a), _a))-4*_b(_a)+1, _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: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 31
dsolve(diff(y(x),x$4)+4*diff(y(x),x$3)+4*diff(y(x),x$2)=1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (2 c_{1} x +2 c_{1} +2 c_{2} \right ) {\mathrm e}^{-2 x}}{8}+\frac {x^{2}}{8}+c_{3} x +c_{4} \]
✓ Solution by Mathematica
Time used: 0.166 (sec). Leaf size: 37
DSolve[y''''[x]+4*y'''[x]+4*y''[x]==1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^2}{8}+c_4 x+\frac {1}{4} e^{-2 x} (c_2 (x+1)+c_1)+c_3 \]