19.52 problem section 9.3, problem 52

Internal problem ID [1549]
Internal file name [OUTPUT/1550_Sunday_June_05_2022_02_21_57_AM_96974772/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 52.
ODE order: 3.
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

[[_3rd_order, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y=12 \,{\mathrm e}^{-x}+9 \cos \left (2 x \right )-13 \sin \left (2 x \right )} \] 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 \[ y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y = 0 \] The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}+3 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3} \] 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 &= x^{2} {\mathrm e}^{-x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y = 12 \,{\mathrm e}^{-x}+9 \cos \left (2 x \right )-13 \sin \left (2 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 12 \,{\mathrm e}^{-x}+9 \cos \left (2 x \right )-13 \sin \left (2 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] Since \(x^{2} {\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{-x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] 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. \[ y_p = A_{1} x^{3} {\mathrm e}^{-x}+A_{2} \cos \left (2 x \right )+A_{3} \sin \left (2 x \right ) \] 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 \[ 6 A_{1} {\mathrm e}^{-x}+2 A_{2} \sin \left (2 x \right )-2 A_{3} \cos \left (2 x \right )-11 A_{2} \cos \left (2 x \right )-11 A_{3} \sin \left (2 x \right ) = 12 \,{\mathrm e}^{-x}+9 \cos \left (2 x \right )-13 \sin \left (2 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 2, A_{2} = -1, A_{3} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 2 x^{3} {\mathrm e}^{-x}-\cos \left (2 x \right )+\sin \left (2 x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3}\right ) + \left (2 x^{3} {\mathrm e}^{-x}-\cos \left (2 x \right )+\sin \left (2 x \right )\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+2 x^{3} {\mathrm e}^{-x}-\cos \left (2 x \right )+\sin \left (2 x \right ) \]

Summary

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

Verification of solutions

\[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+2 x^{3} {\mathrm e}^{-x}-\cos \left (2 x \right )+\sin \left (2 x \right ) \] Verified OK.

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

Solution by Maple

Time used: 0.016 (sec). Leaf size: 35

dsolve(1*diff(y(x),x$3)+3*diff(y(x),x$2)+3*diff(y(x),x)+y(x)=12*exp(-x)+9*cos(2*x)-13*sin(2*x),y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.295 (sec). Leaf size: 46

DSolve[1*y'''[x]+3*y''[x]+3*y'[x]+1*y[x]==12*Exp[-x]+9*Cos[2*x]-13*Sin[2*x],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-x} \left (2 x^3+c_3 x^2+e^x \sin (2 x)-e^x \cos (2 x)+c_2 x+c_1\right ) \]