Internal problem ID [6004]
Internal file name [OUTPUT/5252_Sunday_June_05_2022_03_28_36_PM_69622299/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 2. Linear equations with constant coefficients. Page 93
Problem number: 1(h).
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, _quadrature]]
\[ \boxed {y^{\prime \prime \prime }=x^{2}+{\mathrm e}^{-x} \sin \left (x \right )} \] 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 \[ y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{3} x^{2}+c_{2} x +c_{1} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= x^{2} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime } = x^{2}+{\mathrm e}^{-x} \sin \left (x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x^{2}+{\mathrm e}^{-x} \sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-x} \cos \left (x \right ), {\mathrm e}^{-x} \sin \left (x \right )\}, \{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, x^{2}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} \cos \left (x \right ), {\mathrm e}^{-x} \sin \left (x \right )\}, \{x, x^{2}, x^{3}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} \cos \left (x \right ), {\mathrm e}^{-x} \sin \left (x \right )\}, \{x^{2}, x^{3}, x^{4}\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-x} \cos \left (x \right ), {\mathrm e}^{-x} \sin \left (x \right )\}, \{x^{3}, x^{4}, x^{5}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-x} \cos \left (x \right )+A_{2} {\mathrm e}^{-x} \sin \left (x \right )+A_{3} x^{3}+A_{4} x^{4}+A_{5} x^{5} \] 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 \[ -2 A_{1} {\mathrm e}^{-x} \sin \left (x \right )+2 A_{1} {\mathrm e}^{-x} \cos \left (x \right )+2 A_{2} {\mathrm e}^{-x} \cos \left (x \right )+2 A_{2} {\mathrm e}^{-x} \sin \left (x \right )+6 A_{3}+24 A_{4} x +60 A_{5} x^{2} = x^{2}+{\mathrm e}^{-x} \sin \left (x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{4}}, A_{2} = {\frac {1}{4}}, A_{3} = 0, A_{4} = 0, A_{5} = {\frac {1}{60}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{4}+\frac {{\mathrm e}^{-x} \sin \left (x \right )}{4}+\frac {x^{5}}{60} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{3} x^{2}+c_{2} x +c_{1}\right ) + \left (-\frac {{\mathrm e}^{-x} \cos \left (x \right )}{4}+\frac {{\mathrm e}^{-x} \sin \left (x \right )}{4}+\frac {x^{5}}{60}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x^{2}+c_{2} x +c_{1} -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{4}+\frac {{\mathrm e}^{-x} \sin \left (x \right )}{4}+\frac {x^{5}}{60} \\ \end{align*}
Verification of solutions
\[ y = c_{3} x^{2}+c_{2} x +c_{1} -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{4}+\frac {{\mathrm e}^{-x} \sin \left (x \right )}{4}+\frac {x^{5}}{60} \] Verified OK.
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 33
dsolve(diff(y(x),x$3)=x^2+exp(-x)*sin(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-x} \left (-\cos \left (x \right )+\sin \left (x \right )\right )}{4}+\frac {x^{5}}{60}+\frac {c_{1} x^{2}}{2}+c_{2} x +c_{3} \]
✓ Solution by Mathematica
Time used: 0.114 (sec). Leaf size: 47
DSolve[y'''[x]==x^2+Exp[-x]*Sin[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^5}{60}+c_3 x^2+\frac {1}{4} e^{-x} \sin (x)-\frac {1}{4} e^{-x} \cos (x)+c_2 x+c_1 \]