Internal problem ID [15352]
Internal file name [OUTPUT/15353_Wednesday_May_08_2024_03_56_47_PM_31495563/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.3 Nonhomogeneous linear equations with
constant coefficients. Superposition principle. Exercises page 137
Problem number: 584.
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]]
\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime }=2 x +{\mathrm e}^{x}} \] 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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}+\lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 1\\ \lambda _3 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{x} x \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime } = 2 x +{\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 x +{\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x}\}, \{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{x} x, {\mathrm e}^{x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x}\}, \{x, x^{2}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x\}, \{x, x^{2}\}] \] Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x^{2}\}, \{x, x^{2}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{x} x^{2}+A_{2} x +A_{3} x^{2} \] 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 \[ 2 A_{1} {\mathrm e}^{x}-4 A_{3}+A_{2}+2 A_{3} x = 2 x +{\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{2}}, A_{2} = 4, A_{3} = 1\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{x} x^{2}}{2}+4 x +x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3}\right ) + \left (\frac {{\mathrm e}^{x} x^{2}}{2}+4 x +x^{2}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} +\frac {{\mathrm e}^{x} x^{2}}{2}+4 x +x^{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} +\frac {{\mathrm e}^{x} x^{2}}{2}+4 x +x^{2} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} +\frac {{\mathrm e}^{x} x^{2}}{2}+4 x +x^{2} \] 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] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = 2*(diff(_b(_a), _a))-_b(_a)+2*_a+exp(_a), _b(_a)` *** Sublevel 2 ** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 34
dsolve(diff(y(x),x$3)-2*diff(y(x),x$2)+diff(y(x),x)=2*x+exp(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{2}+\left (2 c_{1} -2\right ) x -2 c_{1} +2 c_{2} +2\right ) {\mathrm e}^{x}}{2}+x^{2}+4 x +c_{3} \]
✓ Solution by Mathematica
Time used: 0.268 (sec). Leaf size: 39
DSolve[y'''[x]-2*y''[x]+y'[x]==2*x+Exp[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^2+e^x \left (\frac {x^2}{2}+(-1+c_2) x+1+c_1-c_2\right )+4 x+c_3 \]