problem section 9.3, problem 45

Internal problem ID [1542]
Internal ﬁle name [OUTPUT/1543_Sunday_June_05_2022_02_21_34_AM_62348847/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 45.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+8 y^{\prime \prime }+8 y^{\prime }+4 y=-2 \,{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right )} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+8 y^{\prime \prime }+8 y^{\prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+8 \lambda ^{2}+8 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1-i\\ \lambda _2 &= -1+i\\ \lambda _3 &= -1-i\\ \lambda _4 &= -1+i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{\left (-1+i\right ) x} c_{1} +x \,{\mathrm e}^{\left (-1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1-i\right ) x} c_{3} +x \,{\mathrm e}^{\left (-1-i\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\left (-1+i\right ) x} \\ y_2 &= x \,{\mathrm e}^{\left (-1+i\right ) x} \\ y_3 &= {\mathrm e}^{\left (-1-i\right ) x} \\ y_4 &= x \,{\mathrm e}^{\left (-1-i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+8 y^{\prime \prime }+8 y^{\prime }+4 y = -2 \,{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ -2 \,{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x} \cos \left (x \right ), {\mathrm e}^{x} \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{\left (-1-i\right ) x}, x \,{\mathrm e}^{\left (-1+i\right ) x}, {\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{x} \cos \left (x \right )+A_{2} {\mathrm e}^{x} \sin \left (x \right ) \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ -32 A_{1} {\mathrm e}^{x} \sin \left (x \right )+32 A_{2} {\mathrm e}^{x} \cos \left (x \right ) = -2 \,{\mathrm e}^{x} \left (\cos \left (x \right )-\sin \left (x \right )\right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{16}}, A_{2} = -{\frac {1}{16}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{x} \cos \left (x \right )}{16}-\frac {{\mathrm e}^{x} \sin \left (x \right )}{16} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\left (-1+i\right ) x} c_{1} +x \,{\mathrm e}^{\left (-1+i\right ) x} c_{2} +{\mathrm e}^{\left (-1-i\right ) x} c_{3} +x \,{\mathrm e}^{\left (-1-i\right ) x} c_{4}\right ) + \left (-\frac {{\mathrm e}^{x} \cos \left (x \right )}{16}-\frac {{\mathrm e}^{x} \sin \left (x \right )}{16}\right ) \\ \end{align*} Which simpliﬁes to \[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{\left (-1-i\right ) x}+{\mathrm e}^{\left (-1+i\right ) x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{x} \cos \left (x \right )}{16}-\frac {{\mathrm e}^{x} \sin \left (x \right )}{16} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \right ) {\mathrm e}^{\left (-1-i\right ) x}+{\mathrm e}^{\left (-1+i\right ) x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{x} \cos \left (x \right )}{16}-\frac {{\mathrm e}^{x} \sin \left (x \right )}{16} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{\left (-1-i\right ) x}+{\mathrm e}^{\left (-1+i\right ) x} \left (c_{2} x +c_{1} \right )-\frac {{\mathrm e}^{x} \cos \left (x \right )}{16}-\frac {{\mathrm e}^{x} \sin \left (x \right )}{16} \] 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`

\[ y \left (x \right ) = \left (\left (c_{3} x +c_{1} \right ) \cos \left (x \right )+\sin \left (x \right ) \left (c_{4} x +c_{2} \right )\right ) {\mathrm e}^{-x}-\frac {\left (\sin \left (x \right )+\cos \left (x \right )\right ) {\mathrm e}^{x}}{16} \]

\[ y(x)\to \frac {1}{16} e^{-x} \left (\left (-e^{2 x}+16 (c_4 x+c_3)\right ) \cos (x)-\left (e^{2 x}-16 (c_2 x+c_1)\right ) \sin (x)\right ) \]

19.45 problem section 9.3, problem 45