14.23 problem 23

Internal problem ID [2224]
Internal file name [OUTPUT/2224_Monday_February_26_2024_09_18_32_AM_49298837/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 23, page 106
Problem number: 23.
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, _missing_y]]

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

Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{4 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-4 y^{\prime \prime } = {\mathrm e}^{2 x} \left (x -3\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \left (x -3\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{2 x}, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{4 x}\} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} x \,{\mathrm e}^{2 x}+A_{2} {\mathrm e}^{2 x} \] 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 \[ -4 A_{1} {\mathrm e}^{2 x}-8 A_{1} x \,{\mathrm e}^{2 x}-8 A_{2} {\mathrm e}^{2 x} = {\mathrm e}^{2 x} \left (x -3\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{8}}, A_{2} = {\frac {7}{16}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{2 x}}{8}+\frac {7 \,{\mathrm e}^{2 x}}{16} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3}\right ) + \left (-\frac {x \,{\mathrm e}^{2 x}}{8}+\frac {7 \,{\mathrm e}^{2 x}}{16}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} -\frac {x \,{\mathrm e}^{2 x}}{8}+\frac {7 \,{\mathrm e}^{2 x}}{16} \\ \end{align*}

Verification of solutions

\[ y = c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} -\frac {x \,{\mathrm e}^{2 x}}{8}+\frac {7 \,{\mathrm e}^{2 x}}{16} \] 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) = exp(2*_a)*_a-3*exp(2*_a)+4*_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`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 27

dsolve(diff(y(x),x$3)-4*diff(y(x),x$2)=exp(2*x)*(x-3),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (-2 x +7\right ) {\mathrm e}^{2 x}}{16}+c_{2} x +\frac {{\mathrm e}^{4 x} c_{1}}{16}+c_{3} \]

Solution by Mathematica

Time used: 0.268 (sec). Leaf size: 34

DSolve[y'''[x]-4*y''[x]==Exp[2*x]*(x-3),y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{16} e^{2 x} \left (-2 x+c_1 e^{2 x}+7\right )+c_3 x+c_2 \]