Internal problem ID [12201]
Internal file name [OUTPUT/10854_Thursday_September_21_2023_05_48_07_AM_3695680/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 54.
ODE order: 6.
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]]
\[ \boxed {y^{\left (6\right )}+2 y^{\prime \prime \prime \prime }+y^{\prime \prime }=x +{\mathrm e}^{x}} \] 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 \[ y^{\left (6\right )}+2 y^{\prime \prime \prime \prime }+y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{6}+2 \lambda ^{4}+\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= i\\ \lambda _4 &= -i\\ \lambda _5 &= i\\ \lambda _6 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{-i x} c_{3} +x \,{\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{5} +x \,{\mathrm e}^{i x} c_{6} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{-i x} \\ y_4 &= x \,{\mathrm e}^{-i x} \\ y_5 &= {\mathrm e}^{i x} \\ y_6 &= x \,{\mathrm e}^{i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (6\right )}+2 y^{\prime \prime \prime \prime }+y^{\prime \prime } = x +{\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x +{\mathrm e}^{x} \] 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 \[ \{1, x, x \,{\mathrm e}^{i x}, x \,{\mathrm e}^{-i x}, {\mathrm e}^{i x}, {\mathrm e}^{-i x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x}\}, \{x, x^{2}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x}\}, \{x^{2}, x^{3}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{x}+A_{2} x^{2}+A_{3} x^{3} \] 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 \[ 4 A_{1} {\mathrm e}^{x}+2 A_{2}+6 A_{3} x = x +{\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{4}}, A_{2} = 0, A_{3} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{x}}{4}+\frac {x^{3}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{-i x} c_{3} +x \,{\mathrm e}^{-i x} c_{4} +{\mathrm e}^{i x} c_{5} +x \,{\mathrm e}^{i x} c_{6}\right ) + \left (\frac {{\mathrm e}^{x}}{4}+\frac {x^{3}}{6}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-i x}+\left (c_{6} x +c_{5} \right ) {\mathrm e}^{i x}+c_{2} x +c_{1} +\frac {{\mathrm e}^{x}}{4}+\frac {x^{3}}{6} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-i x}+\left (c_{6} x +c_{5} \right ) {\mathrm e}^{i x}+c_{2} x +c_{1} +\frac {{\mathrm e}^{x}}{4}+\frac {x^{3}}{6} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-i x}+\left (c_{6} x +c_{5} \right ) {\mathrm e}^{i x}+c_{2} x +c_{1} +\frac {{\mathrm e}^{x}}{4}+\frac {x^{3}}{6} \] 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: 6; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(diff(diff(_b(_a), _a), _a), _a), _a) = -2*(diff(diff(_b(_a), _a), _a))-_b(_a)+_a+exp(_a) 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] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful <- differential order: 6; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 46
dsolve(diff(y(x),x$6)+2*diff(y(x),x$4)+diff(y(x),x$2)=x+exp(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (-c_{3} x -c_{1} -2 c_{4} \right ) \cos \left (x \right )+\left (-c_{4} x -c_{2} +2 c_{3} \right ) \sin \left (x \right )+\frac {x^{3}}{6}+c_{5} x +c_{6} +\frac {{\mathrm e}^{x}}{4} \]
✓ Solution by Mathematica
Time used: 0.61 (sec). Leaf size: 58
DSolve[y''''''[x]+2*y''''[x]+y''[x]==x+Exp[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^3}{6}+\frac {e^x}{4}+c_6 x-(c_2 x+c_1+2 c_4) \cos (x)+(-c_4 x+2 c_2-c_3) \sin (x)+c_5 \]