problem section 9.3, problem 21

Internal problem ID [1518]
Internal ﬁle name [OUTPUT/1519_Sunday_June_05_2022_02_20_26_AM_55030780/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 21.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {2 y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime }-y=3 \,{\mathrm e}^{-\frac {x}{2}} \left (1-6 x \right )} \] 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 \[ 2 y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime }-y = 0 \] The characteristic equation is \[ 2 \lambda ^{4}+\lambda ^{3}-2 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -\frac {1}{2}-\frac {i \sqrt {3}}{2}\\ \lambda _4 &= -\frac {1}{2}+\frac {i \sqrt {3}}{2} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +{\mathrm e}^{-\frac {x}{2}} c_{3} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} \\ y_3 &= {\mathrm e}^{-\frac {x}{2}} \\ y_4 &= {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ 2 y^{\prime \prime \prime \prime }+y^{\prime \prime \prime }-2 y^{\prime }-y = 3 \,{\mathrm e}^{-\frac {x}{2}} \left (1-6 x \right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ 3 \,{\mathrm e}^{-\frac {x}{2}} \left (1-6 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ \left [\left \{x \,{\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{-\frac {x}{2}}\right \}\right ] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{x}, {\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x}, {\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x}, {\mathrm e}^{-\frac {x}{2}}\right \} \] Since \({\mathrm e}^{-\frac {x}{2}}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ \left [\left \{x \,{\mathrm e}^{-\frac {x}{2}}, x^{2} {\mathrm e}^{-\frac {x}{2}}\right \}\right ] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{-\frac {x}{2}}+A_{2} x^{2} {\mathrm e}^{-\frac {x}{2}} \] 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 \[ -\frac {9 A_{1} {\mathrm e}^{-\frac {x}{2}}}{4}+3 A_{2} {\mathrm e}^{-\frac {x}{2}}-\frac {9 A_{2} x \,{\mathrm e}^{-\frac {x}{2}}}{2} = 3 \,{\mathrm e}^{-\frac {x}{2}} \left (1-6 x \right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ [A_{1} = 4, A_{2} = 4] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 4 x \,{\mathrm e}^{-\frac {x}{2}}+4 x^{2} {\mathrm e}^{-\frac {x}{2}} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x} c_{2} +{\mathrm e}^{-\frac {x}{2}} c_{3} +{\mathrm e}^{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x} c_{4}\right ) + \left (4 x \,{\mathrm e}^{-\frac {x}{2}}+4 x^{2} {\mathrm e}^{-\frac {x}{2}}\right ) \\ \end{align*}

Summary

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

Veriﬁcation of solutions

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

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

\[ y(x)\to e^{-x/2} \left (4 x^2+4 x+c_4 e^{3 x/2}+c_2 \cos \left (\frac {\sqrt {3} x}{2}\right )+c_1 \sin \left (\frac {\sqrt {3} x}{2}\right )-8+c_3\right ) \]

19.21 problem section 9.3, problem 21