Internal problem ID [11791]
Internal file name [OUTPUT/11801_Thursday_April_11_2024_08_49_47_PM_42252830/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: 18.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+2 y^{\prime \prime }=3 \,{\mathrm e}^{-x}+6 \,{\mathrm e}^{2 x}-6 x} \] 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 \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+2 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}-3 \lambda ^{3}+2 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 2\\ \lambda _4 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{x} \\ y_4 &= {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+2 y^{\prime \prime } = 3 \,{\mathrm e}^{-x}+6 \,{\mathrm e}^{2 x}-6 x \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 3 \,{\mathrm e}^{-x}+6 \,{\mathrm e}^{2 x}-6 x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x}\}, \{{\mathrm e}^{2 x}\}, \{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{x}, {\mathrm e}^{2 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x}\}, \{{\mathrm e}^{2 x}\}, \{x, x^{2}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x}\}, \{{\mathrm e}^{2 x}\}, \{x^{2}, x^{3}\}] \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x}\}, \{{\mathrm e}^{2 x} x\}, \{x^{2}, x^{3}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-x}+A_{2} {\mathrm e}^{2 x} x +A_{3} x^{2}+A_{4} x^{3} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ 6 A_{1} {\mathrm e}^{-x}+4 A_{2} {\mathrm e}^{2 x}-18 A_{4}+4 A_{3}+12 A_{4} x = 3 \,{\mathrm e}^{-x}+6 \,{\mathrm e}^{2 x}-6 x \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{2}}, A_{2} = {\frac {3}{2}}, A_{3} = -{\frac {9}{4}}, A_{4} = -{\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{-x}}{2}+\frac {3 \,{\mathrm e}^{2 x} x}{2}-\frac {9 x^{2}}{4}-\frac {x^{3}}{2} \] 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}^{2 x} c_{4}\right ) + \left (\frac {{\mathrm e}^{-x}}{2}+\frac {3 \,{\mathrm e}^{2 x} x}{2}-\frac {9 x^{2}}{4}-\frac {x^{3}}{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}^{2 x} c_{4} +\frac {{\mathrm e}^{-x}}{2}+\frac {3 \,{\mathrm e}^{2 x} x}{2}-\frac {9 x^{2}}{4}-\frac {x^{3}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x +c_{1} +c_{3} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{4} +\frac {{\mathrm e}^{-x}}{2}+\frac {3 \,{\mathrm e}^{2 x} x}{2}-\frac {9 x^{2}}{4}-\frac {x^{3}}{2} \] 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) = 3*(diff(_b(_a), _a))-2*_b(_a)+3*exp(-_a)+6*exp(2*_a)-6*_a, _b(_a)` 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)] <- double symmetry of the form [xi=0, eta=F(x)] successful <- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 41
dsolve(diff(y(x),x$4)-3*diff(y(x),x$3)+2*diff(y(x),x$2)=3*exp(-x)+6*exp(2*x)-6*x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (6 x +c_{1} -12\right ) {\mathrm e}^{2 x}}{4}-\frac {x^{3}}{2}-\frac {9 x^{2}}{4}+c_{3} x +c_{2} {\mathrm e}^{x}+c_{4} +\frac {{\mathrm e}^{-x}}{2} \]
✓ Solution by Mathematica
Time used: 0.372 (sec). Leaf size: 54
DSolve[y''''[x]-3*y'''[x]+2*y''[x]==3*Exp[-x]+6*Exp[2*x]-6*x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} \left (-\left ((2 x+9) x^2\right )+2 e^{-x}+4 c_1 e^x+e^{2 x} (6 x-12+c_2)\right )+c_4 x+c_3 \]