Internal problem ID [2242]
Internal file name [OUTPUT/2242_Monday_February_26_2024_09_18_39_AM_47990902/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 24, page 109
Problem number: 10.
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]]
\[ \boxed {y^{\prime \prime \prime }+2 y^{\prime \prime }=x \cos \left (2 x \right )} \] 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 \[ y^{\prime \prime \prime }+2 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3}+2 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -2\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{-2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+2 y^{\prime \prime } = x \cos \left (2 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x \cos \left (2 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \cos \left (2 x \right ), x \sin \left (2 x \right ), \cos \left (2 x \right ), \sin \left (2 x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{-2 x}\} \] 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. \[ y_p = A_{1} x \cos \left (2 x \right )+A_{2} x \sin \left (2 x \right )+A_{3} \cos \left (2 x \right )+A_{4} \sin \left (2 x \right ) \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ -12 A_{1} \cos \left (2 x \right )+8 A_{1} x \sin \left (2 x \right )-12 A_{2} \sin \left (2 x \right )-8 A_{2} x \cos \left (2 x \right )+8 A_{3} \sin \left (2 x \right )-8 A_{4} \cos \left (2 x \right )-8 A_{1} \sin \left (2 x \right )-8 A_{1} x \cos \left (2 x \right )+8 A_{2} \cos \left (2 x \right )-8 A_{2} x \sin \left (2 x \right )-8 A_{3} \cos \left (2 x \right )-8 A_{4} \sin \left (2 x \right ) = x \cos \left (2 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {1}{16}}, A_{2} = -{\frac {1}{16}}, A_{3} = -{\frac {1}{16}}, A_{4} = {\frac {3}{32}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \cos \left (2 x \right )}{16}-\frac {x \sin \left (2 x \right )}{16}-\frac {\cos \left (2 x \right )}{16}+\frac {3 \sin \left (2 x \right )}{32} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3}\right ) + \left (-\frac {x \cos \left (2 x \right )}{16}-\frac {x \sin \left (2 x \right )}{16}-\frac {\cos \left (2 x \right )}{16}+\frac {3 \sin \left (2 x \right )}{32}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} -\frac {x \cos \left (2 x \right )}{16}-\frac {x \sin \left (2 x \right )}{16}-\frac {\cos \left (2 x \right )}{16}+\frac {3 \sin \left (2 x \right )}{32} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x +c_{1} +{\mathrm e}^{-2 x} c_{3} -\frac {x \cos \left (2 x \right )}{16}-\frac {x \sin \left (2 x \right )}{16}-\frac {\cos \left (2 x \right )}{16}+\frac {3 \sin \left (2 x \right )}{32} \] 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) = _a*cos(2*_a)-2*_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: 38
dsolve(diff(y(x),x$3)+2*diff(y(x),x$2)=x*cos(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-2 x -2\right ) \cos \left (2 x \right )}{32}+\frac {\left (3-2 x \right ) \sin \left (2 x \right )}{32}+c_{2} x +\frac {{\mathrm e}^{-2 x} c_{1}}{4}+c_{3} \]
✓ Solution by Mathematica
Time used: 1.447 (sec). Leaf size: 48
DSolve[y'''[x]+2*y''[x]==x*Cos[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_3 x+\frac {1}{16} \left (-x \sin (2 x)-(x+1) \cos (2 x)+4 c_1 e^{-2 x}+3 \sin (x) \cos (x)\right )+c_2 \]