Internal problem ID [5371]
Internal file name [OUTPUT/4862_Sunday_February_04_2024_12_46_34_AM_81323681/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 14. Linear equations with constant coefficients. Supplemetary problems. Page
92
Problem number: 14.
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_x]]
\[ \boxed {y^{\left (5\right )}-4 y^{\prime \prime \prime }=5} \] 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 \[ y^{\left (5\right )}-4 y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{5}-4 \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 2\\ \lambda _5 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=x^{2} c_{3} +c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{4} +{\mathrm e}^{2 x} c_{5} \] 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 &= {\mathrm e}^{-2 x} \\ y_5 &= {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (5\right )}-4 y^{\prime \prime \prime } = 5 \] 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 \[ \{1, x, x^{2}, {\mathrm e}^{-2 x}, {\mathrm e}^{2 x}\} \] 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 \(x^{2}\) 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 \[ -24 A_{1} = 5 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {5}{24}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {5 x^{3}}{24} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x^{2} c_{3} +c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{4} +{\mathrm e}^{2 x} c_{5}\right ) + \left (-\frac {5 x^{3}}{24}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2} c_{3} +c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{4} +{\mathrm e}^{2 x} c_{5} -\frac {5 x^{3}}{24} \\ \end{align*}
Verification of solutions
\[ y = x^{2} c_{3} +c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{4} +{\mathrm e}^{2 x} c_{5} -\frac {5 x^{3}}{24} \] 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) = 4*_b(_a)+5, _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: 0.015 (sec). Leaf size: 34
dsolve(diff(y(x),x$5)-4*diff(y(x),x$3)=5,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {5 x^{3}}{24}+\frac {c_{2} {\mathrm e}^{2 x}}{8}+\frac {c_{3} x^{2}}{2}-\frac {{\mathrm e}^{-2 x} c_{1}}{8}+c_{4} x +c_{5} \]
✓ Solution by Mathematica
Time used: 0.079 (sec). Leaf size: 47
DSolve[y'''''[x]-4*y'''[x]==5,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {5 x^3}{24}+c_5 x^2+c_4 x+\frac {1}{8} c_1 e^{2 x}-\frac {1}{8} c_2 e^{-2 x}+c_3 \]