problem section 9.3, problem 58

Internal problem ID [1555]
Internal ﬁle name [OUTPUT/1556_Sunday_June_05_2022_02_22_15_AM_95328679/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 58.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }+9 y^{\prime \prime }+7 y^{\prime }+2 y={\mathrm e}^{-x} \left (30+24 x \right )-{\mathrm e}^{-2 x}} \] 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 \[ y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }+9 y^{\prime \prime }+7 y^{\prime }+2 y = 0 \] The characteristic equation is \[ \lambda ^{4}+5 \lambda ^{3}+9 \lambda ^{2}+7 \lambda +2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -2\\ \lambda _2 &= -1\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} x \,{\mathrm e}^{-x}+x^{2} {\mathrm e}^{-x} c_{3} +c_{4} {\mathrm e}^{-2 x} \] 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} \\ y_4 &= {\mathrm e}^{-2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+5 y^{\prime \prime \prime }+9 y^{\prime \prime }+7 y^{\prime }+2 y = {\mathrm e}^{-x} \left (30+24 x \right )-{\mathrm e}^{-2 x} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-x} \left (30+24 x \right )-{\mathrm e}^{-2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{-x}, {\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-2 x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x}\}, \{x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x} x^{3}\}] \] Since \(x^{2} {\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x}\}, \{{\mathrm e}^{-x} x^{3}, {\mathrm e}^{-x} x^{4}\}] \] Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{{\mathrm e}^{-x} x^{3}, {\mathrm e}^{-x} x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-2 x} x +A_{2} {\mathrm e}^{-x} x^{3}+A_{3} {\mathrm e}^{-x} x^{4} \] 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 \[ 24 A_{3} {\mathrm e}^{-x} x -A_{1} {\mathrm e}^{-2 x}+6 A_{2} {\mathrm e}^{-x}+24 A_{3} {\mathrm e}^{-x} = {\mathrm e}^{-x} \left (30+24 x \right )-{\mathrm e}^{-2 x} \] Solving for the unknowns by comparing coeﬃcients results in \[ [A_{1} = 1, A_{2} = 1, A_{3} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{-2 x} x +{\mathrm e}^{-x} x^{3}+{\mathrm e}^{-x} x^{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} x \,{\mathrm e}^{-x}+x^{2} {\mathrm e}^{-x} c_{3} +c_{4} {\mathrm e}^{-2 x}\right ) + \left ({\mathrm e}^{-2 x} x +{\mathrm e}^{-x} x^{3}+{\mathrm e}^{-x} x^{4}\right ) \\ \end{align*} Which simpliﬁes to \[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+c_{4} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-x} x^{3}+{\mathrm e}^{-x} x^{4} \]

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 )+c_{4} {\mathrm e}^{-2 x}+{\mathrm e}^{-2 x} x +{\mathrm e}^{-x} x^{3}+{\mathrm e}^{-x} x^{4} \\ \end{align*}

Veriﬁcation of solutions

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

\[ y \left (x \right ) = \left (x^{4}+x^{3}+\left (c_{4} -3\right ) x^{2}+\left (c_{3} +6\right ) x +c_{2} -6\right ) {\mathrm e}^{-x}+{\mathrm e}^{-2 x} \left (x +c_{1} +3\right ) \]

\[ y(x)\to e^{-2 x} \left (e^x \left (x^4+x^3+(-3+c_4) x^2+(6+c_3) x-6+c_2\right )+x+3+c_1\right ) \]

19.58 problem section 9.3, problem 58