Internal problem ID [1562]
Internal file name [OUTPUT/1563_Sunday_June_05_2022_02_22_32_AM_42329436/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 65.
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }-2 y^{\prime \prime }+y=-{\mathrm e}^{-x} \left (3 x^{2}-9 x +4\right )} \] 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 \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime }+y = 0 \] The characteristic equation is \[ \lambda ^{4}-2 \lambda ^{2}+1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 1\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{-x} x \\ y_3 &= {\mathrm e}^{x} \\ y_4 &= x \,{\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime }+y = -{\mathrm e}^{-x} \left (3 x^{2}-9 x +4\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -{\mathrm e}^{-x} \left (3 x^{2}-9 x +4\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x} x, {\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{x}, {\mathrm e}^{-x} x, {\mathrm e}^{x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}, {\mathrm e}^{-x} x\}] \] Since \({\mathrm e}^{-x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}, x^{4} {\mathrm e}^{-x}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{-x}+A_{2} x^{3} {\mathrm e}^{-x}+A_{3} x^{4} {\mathrm e}^{-x} \] 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 \[ 8 A_{1} {\mathrm e}^{-x}-24 A_{2} {\mathrm e}^{-x}+24 A_{2} x \,{\mathrm e}^{-x}+24 A_{3} {\mathrm e}^{-x}-96 A_{3} x \,{\mathrm e}^{-x}+48 A_{3} x^{2} {\mathrm e}^{-x} = -{\mathrm e}^{-x} \left (3 x^{2}-9 x +4\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{16}}, A_{2} = {\frac {1}{8}}, A_{3} = -{\frac {1}{16}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{2} {\mathrm e}^{-x}}{16}+\frac {x^{3} {\mathrm e}^{-x}}{8}-\frac {x^{4} {\mathrm e}^{-x}}{16} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{x} c_{3} +x \,{\mathrm e}^{x} c_{4}\right ) + \left (\frac {x^{2} {\mathrm e}^{-x}}{16}+\frac {x^{3} {\mathrm e}^{-x}}{8}-\frac {x^{4} {\mathrm e}^{-x}}{16}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (c_{4} x +c_{3} \right )+\frac {x^{2} {\mathrm e}^{-x}}{16}+\frac {x^{3} {\mathrm e}^{-x}}{8}-\frac {x^{4} {\mathrm e}^{-x}}{16} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (c_{4} x +c_{3} \right )+\frac {x^{2} {\mathrm e}^{-x}}{16}+\frac {x^{3} {\mathrm e}^{-x}}{8}-\frac {x^{4} {\mathrm e}^{-x}}{16} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (c_{4} x +c_{3} \right )+\frac {x^{2} {\mathrm e}^{-x}}{16}+\frac {x^{3} {\mathrm e}^{-x}}{8}-\frac {x^{4} {\mathrm e}^{-x}}{16} \] 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] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 40
dsolve(diff(y(x),x$4)-0*diff(y(x),x$3)-2*diff(y(x),x$2)+0*diff(y(x),x)+y(x)=-exp(-x)*(4-9*x+3*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-x^{4}+2 x^{3}+16 c_{4} x +x^{2}+16 c_{2} \right ) {\mathrm e}^{-x}}{16}+{\mathrm e}^{x} \left (c_{3} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.113 (sec). Leaf size: 61
DSolve[y''''[x]-0*y'''[x]-2*y''[x]+0*y'[x]+1*y[x]==-Exp[-x]*(4-9*x+3*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{32} e^{-x} \left (-2 x^4+4 x^3+2 x^2+x \left (32 c_4 e^{2 x}-2+32 c_2\right )+32 c_3 e^{2 x}-3+32 c_1\right ) \]