Internal problem ID [5183]
Internal file name [OUTPUT/4676_Sunday_June_05_2022_03_03_10_PM_88926433/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 11. THE METHOD OF UNDETERMINED COEFFICIENTS. Supplementary
Problems. page 101
Problem number: Problem 11.52.
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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y={\mathrm e}^{x}+1} \] 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 \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{x} x \\ y_3 &= x^{2} {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y = {\mathrm e}^{x}+1 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x}+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}, \{{\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x} x, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{1\}, \{{\mathrm e}^{x} x\}] \] Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{1\}, \{x^{2} {\mathrm e}^{x}\}] \] Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{1\}, \{x^{3} {\mathrm e}^{x}\}] \] 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. \[ y_p = A_{1}+A_{2} x^{3} {\mathrm e}^{x} \] 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 \[ 6 A_{2} {\mathrm e}^{x}-A_{1} = {\mathrm e}^{x}+1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -1, A_{2} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -1+\frac {x^{3} {\mathrm e}^{x}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} c_{3}\right ) + \left (-1+\frac {x^{3} {\mathrm e}^{x}}{6}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )-1+\frac {x^{3} {\mathrm e}^{x}}{6} \]
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 )-1+\frac {x^{3} {\mathrm e}^{x}}{6} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )-1+\frac {x^{3} {\mathrm e}^{x}}{6} \] 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: 0.015 (sec). Leaf size: 27
dsolve(diff(y(x),x$3)-3*diff(y(x),x$2)+3*diff(y(x),x)-y(x)=exp(x)+1,y(x), singsol=all)
\[ y \left (x \right ) = -1+\frac {\left (6 c_{3} x^{2}+x^{3}+6 c_{2} x +6 c_{1} \right ) {\mathrm e}^{x}}{6} \]
✓ Solution by Mathematica
Time used: 0.067 (sec). Leaf size: 31
DSolve[y'''[x]-3*y''[x]+3*y'[x]-y[x]==Exp[x]+1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -1+e^x \left (\frac {x^3}{6}+c_3 x^2+c_2 x+c_1\right ) \]