Internal problem ID [15276]
Internal file name [OUTPUT/15277_Wednesday_May_08_2024_03_54_40_PM_43711821/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: 507.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+6 y^{\prime \prime }+4 y^{\prime }+y=\sin \left (x \right )} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+6 y^{\prime \prime }+4 y^{\prime }+y = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3}+6 \lambda ^{2}+4 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -1\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} x \,{\mathrm e}^{-x}+x^{2} {\mathrm e}^{-x} c_{3} +x^{3} {\mathrm e}^{-x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{-x} x \\ y_3 &= x^{2} {\mathrm e}^{-x} \\ y_4 &= x^{3} {\mathrm e}^{-x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }+6 y^{\prime \prime }+4 y^{\prime }+y = \sin \left (x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ \sin \left (x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (x \right ), \sin \left (x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x^{2} {\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}, {\mathrm e}^{-x} x, {\mathrm e}^{-x}\} \] 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. \[ y_p = A_{1} \cos \left (x \right )+A_{2} \sin \left (x \right ) \] 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 \[ -4 A_{1} \cos \left (x \right )-4 A_{2} \sin \left (x \right ) = \sin \left (x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 0, A_{2} = -{\frac {1}{4}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {\sin \left (x \right )}{4} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} x \,{\mathrm e}^{-x}+x^{2} {\mathrm e}^{-x} c_{3} +x^{3} {\mathrm e}^{-x} c_{4}\right ) + \left (-\frac {\sin \left (x \right )}{4}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )-\frac {\sin \left (x \right )}{4} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )-\frac {\sin \left (x \right )}{4} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-x} \left (c_{4} x^{3}+c_{3} x^{2}+c_{2} x +c_{1} \right )-\frac {\sin \left (x \right )}{4} \] 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: 0.0 (sec). Leaf size: 29
dsolve(diff(y(x),x$4)+4*diff(y(x),x$3)+6*diff(y(x),x$2)+4*diff(y(x),x)+y(x)=sin(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{3} x^{3}+c_{2} x^{2}+c_{4} x +c_{1} \right ) {\mathrm e}^{-x}-\frac {\sin \left (x \right )}{4} \]
✓ Solution by Mathematica
Time used: 0.102 (sec). Leaf size: 35
DSolve[y''''[x]+4*y'''[x]+6*y''[x]+4*y'[x]+y[x]==Sin[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {\sin (x)}{4}+e^{-x} (x (x (c_4 x+c_3)+c_2)+c_1) \]