Internal problem ID [15273]
Internal file name [OUTPUT/15274_Wednesday_May_08_2024_03_54_36_PM_62930676/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. Trial and error method. Exercises page 132
Problem number: 504.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime }+4 y=\sin \left (2 x \right ) x} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{2}+4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= i \sqrt {2}\\ \lambda _2 &= -i \sqrt {2}\\ \lambda _3 &= i \sqrt {2}\\ \lambda _4 &= -i \sqrt {2} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-i \sqrt {2}\, x} c_{1} +x \,{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3} +x \,{\mathrm e}^{i \sqrt {2}\, x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-i \sqrt {2}\, x} \\ y_2 &= x \,{\mathrm e}^{-i \sqrt {2}\, x} \\ y_3 &= {\mathrm e}^{i \sqrt {2}\, x} \\ y_4 &= x \,{\mathrm e}^{i \sqrt {2}\, x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime }+4 y = \sin \left (2 x \right ) x \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (2 x \right ) x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (2 x \right ) x, \sin \left (2 x \right ) x, \cos \left (2 x \right ), \sin \left (2 x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{x \,{\mathrm e}^{i \sqrt {2}\, x}, x \,{\mathrm e}^{-i \sqrt {2}\, x}, {\mathrm e}^{i \sqrt {2}\, x}, {\mathrm e}^{-i \sqrt {2}\, x}\right \} \] 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. \[ y_p = A_{1} \cos \left (2 x \right ) x +A_{2} \sin \left (2 x \right ) x +A_{3} \cos \left (2 x \right )+A_{4} \sin \left (2 x \right ) \] 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 \[ 4 A_{1} \cos \left (2 x \right ) x +16 A_{1} \sin \left (2 x \right )+4 A_{2} \sin \left (2 x \right ) x -16 A_{2} \cos \left (2 x \right )+4 A_{3} \cos \left (2 x \right )+4 A_{4} \sin \left (2 x \right ) = \sin \left (2 x \right ) x \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{4}}, A_{3} = 1, A_{4} = 0\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {\sin \left (2 x \right ) x}{4}+\cos \left (2 x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-i \sqrt {2}\, x} c_{1} +x \,{\mathrm e}^{-i \sqrt {2}\, x} c_{2} +{\mathrm e}^{i \sqrt {2}\, x} c_{3} +x \,{\mathrm e}^{i \sqrt {2}\, x} c_{4}\right ) + \left (\frac {\sin \left (2 x \right ) x}{4}+\cos \left (2 x \right )\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-i \sqrt {2}\, x}+\left (c_{4} x +c_{3} \right ) {\mathrm e}^{i \sqrt {2}\, x}+\frac {\sin \left (2 x \right ) x}{4}+\cos \left (2 x \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-i \sqrt {2}\, x}+\left (c_{4} x +c_{3} \right ) {\mathrm e}^{i \sqrt {2}\, x}+\frac {\sin \left (2 x \right ) x}{4}+\cos \left (2 x \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-i \sqrt {2}\, x}+\left (c_{4} x +c_{3} \right ) {\mathrm e}^{i \sqrt {2}\, x}+\frac {\sin \left (2 x \right ) x}{4}+\cos \left (2 x \right ) \] 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: 40
dsolve(diff(y(x),x$4)+4*diff(y(x),x$2)+4*y(x)=x*sin(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{3} x +c_{1} \right ) \cos \left (\sqrt {2}\, x \right )+\left (c_{4} x +c_{2} \right ) \sin \left (\sqrt {2}\, x \right )+\frac {x \sin \left (2 x \right )}{4}+\cos \left (2 x \right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 58
DSolve[y''''[x]+4*y''[x]+4*y[x]==x*Sin[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} x \sin (2 x)+\cos (2 x)+(c_2 x+c_1) \cos \left (\sqrt {2} x\right )+c_3 \sin \left (\sqrt {2} x\right )+c_4 x \sin \left (\sqrt {2} x\right ) \]