problem section 9.3, problem 28

Internal problem ID [1525]
Internal ﬁle name [OUTPUT/1526_Sunday_June_05_2022_02_20_40_AM_76968418/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 28.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }-7 y^{\prime \prime \prime }+18 y^{\prime \prime }-20 y^{\prime }+8 y={\mathrm e}^{2 x} \left (-5 x^{2}-8 x +3\right )} \] 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 \[ y^{\prime \prime \prime \prime }-7 y^{\prime \prime \prime }+18 y^{\prime \prime }-20 y^{\prime }+8 y = 0 \] The characteristic equation is \[ \lambda ^{4}-7 \lambda ^{3}+18 \lambda ^{2}-20 \lambda +8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= 2\\ \lambda _4 &= 2 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+x \,{\mathrm e}^{2 x} c_{3} +x^{2} {\mathrm e}^{2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{2 x} \\ y_3 &= x \,{\mathrm e}^{2 x} \\ y_4 &= x^{2} {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-7 y^{\prime \prime \prime }+18 y^{\prime \prime }-20 y^{\prime }+8 y = {\mathrm e}^{2 x} \left (-5 x^{2}-8 x +3\right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \left (-5 x^{2}-8 x +3\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{2 x}, x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{3}\}] \] Since \(x \,{\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{3}, {\mathrm e}^{2 x} x^{4}\}] \] Since \(x^{2} {\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{5} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{3}, {\mathrm e}^{2 x} x^{4}\}] \] 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. \[ y_p = A_{1} x^{5} {\mathrm e}^{2 x}+A_{2} {\mathrm e}^{2 x} x^{3}+A_{3} {\mathrm e}^{2 x} x^{4} \] 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 \[ 60 A_{1} x^{2} {\mathrm e}^{2 x}+24 A_{3} {\mathrm e}^{2 x} x +120 A_{1} x \,{\mathrm e}^{2 x}+6 A_{2} {\mathrm e}^{2 x}+24 A_{3} {\mathrm e}^{2 x} = {\mathrm e}^{2 x} \left (-5 x^{2}-8 x +3\right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{12}}, A_{2} = {\frac {1}{6}}, A_{3} = {\frac {1}{12}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x^{5} {\mathrm e}^{2 x}}{12}+\frac {{\mathrm e}^{2 x} x^{3}}{6}+\frac {{\mathrm e}^{2 x} x^{4}}{12} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{2 x}+x \,{\mathrm e}^{2 x} c_{3} +x^{2} {\mathrm e}^{2 x} c_{4}\right ) + \left (-\frac {x^{5} {\mathrm e}^{2 x}}{12}+\frac {{\mathrm e}^{2 x} x^{3}}{6}+\frac {{\mathrm e}^{2 x} x^{4}}{12}\right ) \\ \end{align*} Which simpliﬁes to \[ y = \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{2 x}+c_{1} {\mathrm e}^{x}-\frac {x^{5} {\mathrm e}^{2 x}}{12}+\frac {{\mathrm e}^{2 x} x^{3}}{6}+\frac {{\mathrm e}^{2 x} x^{4}}{12} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{2 x}+c_{1} {\mathrm e}^{x}-\frac {x^{5} {\mathrm e}^{2 x}}{12}+\frac {{\mathrm e}^{2 x} x^{3}}{6}+\frac {{\mathrm e}^{2 x} x^{4}}{12} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{2 x}+c_{1} {\mathrm e}^{x}-\frac {x^{5} {\mathrm e}^{2 x}}{12}+\frac {{\mathrm e}^{2 x} x^{3}}{6}+\frac {{\mathrm e}^{2 x} x^{4}}{12} \] 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`

\[ y \left (x \right ) = -\frac {{\mathrm e}^{x} \left (\left (x^{5}-x^{4}-12 c_{4} x^{2}-2 x^{3}-12 c_{3} x -12 c_{2} \right ) {\mathrm e}^{x}-12 c_{1} \right )}{12} \]

\[ y(x)\to \frac {1}{12} e^{2 x} \left (-x^5+x^4+2 x^3+6 (-1+2 c_4) x^2+12 (1+c_3) x+12 (-1+c_2)\right )+c_1 e^x \]

19.28 problem section 9.3, problem 28