Internal problem ID [1561]
Internal file name [OUTPUT/1562_Sunday_June_05_2022_02_22_30_AM_22651286/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 64.
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y=\left (x +1\right ) {\mathrm e}^{x}} \] 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 \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y = 0 \] The characteristic equation is \[ \lambda ^{3}-3 \lambda ^{2}+3 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +x^{2} {\mathrm e}^{x} c_{3} \] 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} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y = \left (x +1\right ) {\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \left (x +1\right ) {\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] Since \(x \,{\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x^{3}\}] \] Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x^{3}, {\mathrm e}^{x} x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{x} x^{3}+A_{2} {\mathrm e}^{x} x^{4} \] 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 \[ 6 A_{1} {\mathrm e}^{x}+24 A_{2} {\mathrm e}^{x} x = \left (x +1\right ) {\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = {\frac {1}{24}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{24} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{x} c_{2} x +x^{2} {\mathrm e}^{x} c_{3}\right ) + \left (\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{24}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{24} \]
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 )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{24} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{24} \] Verified OK.
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 33
dsolve(diff(y(x),x$3)-3*diff(y(x),x$2)+3*diff(y(x),x)-y(x)=exp(x)*(1+x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{x} \left (x^{4}+4 x^{3}+\left (24 c_{3} +3\right ) x^{2}+24 c_{2} x +24 c_{1} \right )}{24} \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 37
DSolve[y'''[x]-3*y''[x]+3*y'[x]-1*y[x]==Exp[x]*(1+x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{24} e^x \left (x^4+4 x^3+24 c_3 x^2+24 c_2 x+24 c_1\right ) \]