Internal problem ID [1537]
Internal file name [OUTPUT/1538_Sunday_June_05_2022_02_21_14_AM_13504486/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 40.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+13 y^{\prime \prime }+12 y^{\prime }+4 y={\mathrm e}^{-x} \left (\left (4-x \right ) \cos \left (x \right )-\left (5+x \right ) \sin \left (x \right )\right )} \] 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 \[ y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+13 y^{\prime \prime }+12 y^{\prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+6 \lambda ^{3}+13 \lambda ^{2}+12 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -2\\ \lambda _2 &= -2\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{-2 x} c_{3} +x \,{\mathrm e}^{-2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= x \,{\mathrm e}^{-x} \\ y_3 &= {\mathrm e}^{-2 x} \\ y_4 &= x \,{\mathrm e}^{-2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+6 y^{\prime \prime \prime }+13 y^{\prime \prime }+12 y^{\prime }+4 y = {\mathrm e}^{-x} \left (\left (4-x \right ) \cos \left (x \right )-\left (5+x \right ) \sin \left (x \right )\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-x} \left (\left (4-x \right ) \cos \left (x \right )-\left (5+x \right ) \sin \left (x \right )\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x} \cos \left (x \right ), {\mathrm e}^{-x} \sin \left (x \right ), {\mathrm e}^{-x} \cos \left (x \right ) x, {\mathrm e}^{-x} \sin \left (x \right ) x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-2 x}, x \,{\mathrm e}^{-x}, {\mathrm e}^{-2 x}, {\mathrm e}^{-x}\} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} {\mathrm e}^{-x} \cos \left (x \right )+A_{2} {\mathrm e}^{-x} \sin \left (x \right )+A_{3} {\mathrm e}^{-x} \cos \left (x \right ) x +A_{4} {\mathrm e}^{-x} \sin \left (x \right ) x \] 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 \[ -6 A_{3} {\mathrm e}^{-x} \cos \left (x \right )+2 A_{3} {\mathrm e}^{-x} \sin \left (x \right ) x -2 A_{4} {\mathrm e}^{-x} \cos \left (x \right ) x -6 A_{4} {\mathrm e}^{-x} \sin \left (x \right )+2 A_{3} {\mathrm e}^{-x} \sin \left (x \right )-2 A_{4} {\mathrm e}^{-x} \cos \left (x \right )+2 A_{1} {\mathrm e}^{-x} \sin \left (x \right )-2 A_{2} {\mathrm e}^{-x} \cos \left (x \right ) = {\mathrm e}^{-x} \left (\left (4-x \right ) \cos \left (x \right )-\left (5+x \right ) \sin \left (x \right )\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{2}}, A_{2} = -1, A_{3} = -{\frac {1}{2}}, A_{4} = {\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2}-{\mathrm e}^{-x} \sin \left (x \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right ) x}{2}+\frac {{\mathrm e}^{-x} \sin \left (x \right ) x}{2} \] 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}^{-2 x} c_{3} +x \,{\mathrm e}^{-2 x} c_{4}\right ) + \left (-\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2}-{\mathrm e}^{-x} \sin \left (x \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right ) x}{2}+\frac {{\mathrm e}^{-x} \sin \left (x \right ) x}{2}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2}-{\mathrm e}^{-x} \sin \left (x \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right ) x}{2}+\frac {{\mathrm e}^{-x} \sin \left (x \right ) x}{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2}-{\mathrm e}^{-x} \sin \left (x \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right ) x}{2}+\frac {{\mathrm e}^{-x} \sin \left (x \right ) x}{2} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{-2 x}+{\mathrm e}^{-x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2}-{\mathrm e}^{-x} \sin \left (x \right )-\frac {{\mathrm e}^{-x} \cos \left (x \right ) x}{2}+\frac {{\mathrm e}^{-x} \sin \left (x \right ) x}{2} \] 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] 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: 43
dsolve(1*diff(y(x),x$4)+6*diff(y(x),x$3)+13*diff(y(x),x$2)+12*diff(y(x),x)+4*y(x)=exp(-1*x)*((4-x)*cos(x)-(5+x)*sin(x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (-x -1\right ) \cos \left (x \right )+\sin \left (x \right ) \left (-2+x \right )+2 c_{3} x +2 c_{2} \right ) {\mathrm e}^{-x}}{2}+{\mathrm e}^{-2 x} \left (c_{4} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.022 (sec). Leaf size: 56
DSolve[1*y''''[x]+6*y'''[x]+13*y''[x]+12*y'[x]+4*y[x]==Exp[-1*x]*((4-x)*Cos[x]-(5+x)*Sin[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} e^{-2 x} \left (e^x (x-2) \sin (x)-e^x (x+1) \cos (x)+2 \left (c_2 x+c_3 e^x+c_4 e^x x+c_1\right )\right ) \]