Internal problem ID [5395]
Internal file name [OUTPUT/4886_Sunday_February_04_2024_12_46_49_AM_13873342/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 16. Linear equations with constant coefficients (Short methods). Supplemetary
problems. Page 107
Problem number: 29.
ODE order: 4.
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
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }-y=\sin \left (2 x \right )} \] 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 \[ y^{\prime \prime \prime \prime }-y = 0 \] The characteristic equation is \[ \lambda ^{4}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= i\\ \lambda _4 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} +{\mathrm e}^{-i x} c_{3} +{\mathrm e}^{i x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{-i x} \\ y_4 &= {\mathrm e}^{i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-y = \sin \left (2 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (2 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{i x}, {\mathrm e}^{-x}, {\mathrm e}^{-i x}\} \] 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. \[ y_p = A_{1} \cos \left (2 x \right )+A_{2} \sin \left (2 x \right ) \] 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 \[ 15 A_{1} \cos \left (2 x \right )+15 A_{2} \sin \left (2 x \right ) = \sin \left (2 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{15}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {\sin \left (2 x \right )}{15} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} +{\mathrm e}^{-i x} c_{3} +{\mathrm e}^{i x} c_{4}\right ) + \left (\frac {\sin \left (2 x \right )}{15}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} +{\mathrm e}^{-i x} c_{3} +{\mathrm e}^{i x} c_{4} +\frac {\sin \left (2 x \right )}{15} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x}+{\mathrm e}^{x} c_{2} +{\mathrm e}^{-i x} c_{3} +{\mathrm e}^{i x} c_{4} +\frac {\sin \left (2 x \right )}{15} \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 29
dsolve(diff(y(x),x$4)-y(x)=sin(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sin \left (2 x \right )}{15}+\cos \left (x \right ) c_{1} +{\mathrm e}^{x} c_{2} +c_{3} \sin \left (x \right )+c_{4} {\mathrm e}^{-x} \]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 37
DSolve[y''''[x]-y[x]==Sin[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 e^x+c_3 e^{-x}+c_4 \sin (x)+\cos (x) \left (\frac {2 \sin (x)}{15}+c_2\right ) \]