problem 24

Internal problem ID [2163]
Internal ﬁle name [OUTPUT/2163_Monday_February_26_2024_09_17_58_AM_88498068/index.tex]

Book: Diﬀerential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 19, page 86
Problem number: 24.
ODE order: 3.
ODE degree: 1.

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

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

Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{-2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+2 y^{\prime \prime } = \left (2 x^{2}+x \right ) {\mathrm e}^{-2 x}+5 \cos \left (3 x \right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ \left (2 x^{2}+x \right ) {\mathrm e}^{-2 x}+5 \cos \left (3 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (3 x \right ), \sin \left (3 x \right )\}, \{x^{2} {\mathrm e}^{-2 x}, {\mathrm e}^{-2 x} x, {\mathrm e}^{-2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{-2 x}\} \] Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{\cos \left (3 x \right ), \sin \left (3 x \right )\}, \{x^{2} {\mathrm e}^{-2 x}, x^{3} {\mathrm e}^{-2 x}, {\mathrm e}^{-2 x} x\}] \] 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. \[ y_p = A_{1} \cos \left (3 x \right )+A_{2} \sin \left (3 x \right )+A_{3} x^{2} {\mathrm e}^{-2 x}+A_{4} x^{3} {\mathrm e}^{-2 x}+A_{5} {\mathrm e}^{-2 x} x \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) 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 \[ 27 A_{1} \sin \left (3 x \right )-27 A_{2} \cos \left (3 x \right )-8 A_{3} {\mathrm e}^{-2 x}+8 A_{3} x \,{\mathrm e}^{-2 x}+6 A_{4} {\mathrm e}^{-2 x}-24 A_{4} x \,{\mathrm e}^{-2 x}+12 A_{4} x^{2} {\mathrm e}^{-2 x}+4 A_{5} {\mathrm e}^{-2 x}-18 A_{1} \cos \left (3 x \right )-18 A_{2} \sin \left (3 x \right ) = \left (2 x^{2}+x \right ) {\mathrm e}^{-2 x}+5 \cos \left (3 x \right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {10}{117}}, A_{2} = -{\frac {5}{39}}, A_{3} = {\frac {5}{8}}, A_{4} = {\frac {1}{6}}, A_{5} = 1\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {10 \cos \left (3 x \right )}{117}-\frac {5 \sin \left (3 x \right )}{39}+\frac {5 x^{2} {\mathrm e}^{-2 x}}{8}+\frac {x^{3} {\mathrm e}^{-2 x}}{6}+{\mathrm e}^{-2 x} x \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3}\right ) + \left (-\frac {10 \cos \left (3 x \right )}{117}-\frac {5 \sin \left (3 x \right )}{39}+\frac {5 x^{2} {\mathrm e}^{-2 x}}{8}+\frac {x^{3} {\mathrm e}^{-2 x}}{6}+{\mathrm e}^{-2 x} x\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} -\frac {10 \cos \left (3 x \right )}{117}-\frac {5 \sin \left (3 x \right )}{39}+\frac {5 x^{2} {\mathrm e}^{-2 x}}{8}+\frac {x^{3} {\mathrm e}^{-2 x}}{6}+{\mathrm e}^{-2 x} x \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} -\frac {10 \cos \left (3 x \right )}{117}-\frac {5 \sin \left (3 x \right )}{39}+\frac {5 x^{2} {\mathrm e}^{-2 x}}{8}+\frac {x^{3} {\mathrm e}^{-2 x}}{6}+{\mathrm e}^{-2 x} x \] Veriﬁed OK.

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = 2*exp(-2*_a)*_a^2+exp(-2*_a)*_a+5*cos(3*_a)-2*_b(_a), _b(_a)`   *** Sublevel 2 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful`

\[ y \left (x \right ) = \frac {\left (8 x^{3}+30 x^{2}+12 c_{1} +48 x +33\right ) {\mathrm e}^{-2 x}}{48}+c_{2} x +c_{3} -\frac {10 \cos \left (3 x \right )}{117}-\frac {5 \sin \left (3 x \right )}{39} \]

\[ y(x)\to \frac {1}{48} e^{-2 x} \left (8 x^3+30 x^2+48 x+33+12 c_1\right )-\frac {5}{39} \sin (3 x)-\frac {10}{117} \cos (3 x)+c_3 x+c_2 \]

11.24 problem 24