Internal problem ID [11796]
Internal file name [OUTPUT/11806_Thursday_April_11_2024_08_49_50_PM_86836542/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.3. The method of undetermined coefficients. Exercises page
151
Problem number: 23.
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_y]]
\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime }=18 x^{2}+16 x \,{\mathrm e}^{x}+4 \,{\mathrm e}^{3 x}-9} \] 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 \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+2 \lambda ^{3}-3 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 1\\ \lambda _4 &= -3 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{x} \\ y_4 &= {\mathrm e}^{-3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime } = 18 x^{2}+16 x \,{\mathrm e}^{x}+4 \,{\mathrm e}^{3 x}-9 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 18 x^{2}+16 x \,{\mathrm e}^{x}+4 \,{\mathrm e}^{3 x}-9 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{3 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{x}, {\mathrm e}^{-3 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{3 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}, \{x, x^{2}, x^{3}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{3 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}, \{x^{2}, x^{3}, x^{4}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{3 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x} x^{2}\}, \{x^{2}, x^{3}, x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{3 x}+A_{2} x \,{\mathrm e}^{x}+A_{3} {\mathrm e}^{x} x^{2}+A_{4} x^{2}+A_{5} x^{3}+A_{6} x^{4} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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 \[ 108 A_{1} {\mathrm e}^{3 x}+4 A_{2} {\mathrm e}^{x}+8 A_{3} {\mathrm e}^{x} x +18 A_{3} {\mathrm e}^{x}+24 A_{6}+12 A_{5}+48 A_{6} x -6 A_{4}-18 A_{5} x -36 A_{6} x^{2} = 18 x^{2}+16 x \,{\mathrm e}^{x}+4 \,{\mathrm e}^{3 x}-9 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{27}}, A_{2} = -9, A_{3} = 2, A_{4} = -{\frac {19}{6}}, A_{5} = -{\frac {4}{3}}, A_{6} = -{\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{3 x}}{27}-9 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{2}-\frac {19 x^{2}}{6}-\frac {4 x^{3}}{3}-\frac {x^{4}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{4}\right ) + \left (\frac {{\mathrm e}^{3 x}}{27}-9 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{2}-\frac {19 x^{2}}{6}-\frac {4 x^{3}}{3}-\frac {x^{4}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{4} +\frac {{\mathrm e}^{3 x}}{27}-9 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{2}-\frac {19 x^{2}}{6}-\frac {4 x^{3}}{3}-\frac {x^{4}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{-3 x} c_{4} +\frac {{\mathrm e}^{3 x}}{27}-9 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{2}-\frac {19 x^{2}}{6}-\frac {4 x^{3}}{3}-\frac {x^{4}}{2} \] 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) = 16*_a*exp(_a)+18*_a^2-2*(diff(_b(_a), _a))+3*_b(_a)+4*exp(3*_a)-9, _b 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: 64
dsolve(diff(y(x),x$4)+2*diff(y(x),x$3)-3*diff(y(x),x$2)=18*x^2+16*x*exp(x)+4*exp(3*x)-9,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\left (x^{4}+\frac {8}{3} x^{3}+\frac {19}{3} x^{2}-2 c_{3} x -2 c_{4} \right ) {\mathrm e}^{3 x}+\left (-4 x^{2}+18 x -2 c_{2} -\frac {57}{2}\right ) {\mathrm e}^{4 x}-\frac {2 c_{1}}{9}-\frac {2 \,{\mathrm e}^{6 x}}{27}\right ) {\mathrm e}^{-3 x}}{2} \]
✓ Solution by Mathematica
Time used: 1.232 (sec). Leaf size: 70
DSolve[y''''[x]+2*y'''[x]-3*y''[x]==18*x^2+16*x*Exp[x]+4*Exp[3*x]-9,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {1}{6} \left (3 x^2+8 x+19\right ) x^2+\frac {1}{4} e^x \left (8 x^2-36 x+57+4 c_2\right )+\frac {e^{3 x}}{27}+c_4 x+\frac {1}{9} c_1 e^{-3 x}+c_3 \]