Internal problem ID [15336]
Internal file name [OUTPUT/15337_Wednesday_May_08_2024_03_56_18_PM_45960724/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. Superposition principle. Exercises page 137
Problem number: 568.
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, _missing_y]]
\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime }={\mathrm e}^{x}+3 \sin \left (2 x \right )+1} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -4\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{-4 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= x^{2} \\ y_4 &= {\mathrm e}^{-4 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime \prime } = {\mathrm e}^{x}+3 \sin \left (2 x \right )+1 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x}+3 \sin \left (2 x \right )+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1\}, \{{\mathrm e}^{x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, x^{2}, {\mathrm e}^{-4 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x\}, \{{\mathrm e}^{x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2}\}, \{{\mathrm e}^{x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] Since \(x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3}\}, \{{\mathrm e}^{x}\}, \{\cos \left (2 x \right ), \sin \left (2 x \right )\}] \] 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. \[ y_p = A_{1} x^{3}+A_{2} {\mathrm e}^{x}+A_{3} \cos \left (2 x \right )+A_{4} \sin \left (2 x \right ) \] 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 \[ 5 A_{2} {\mathrm e}^{x}+16 A_{3} \cos \left (2 x \right )+16 A_{4} \sin \left (2 x \right )+24 A_{1}+32 A_{3} \sin \left (2 x \right )-32 A_{4} \cos \left (2 x \right ) = {\mathrm e}^{x}+3 \sin \left (2 x \right )+1 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{24}}, A_{2} = {\frac {1}{5}}, A_{3} = {\frac {3}{40}}, A_{4} = {\frac {3}{80}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{3}}{24}+\frac {{\mathrm e}^{x}}{5}+\frac {3 \cos \left (2 x \right )}{40}+\frac {3 \sin \left (2 x \right )}{80} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{-4 x} c_{4}\right ) + \left (\frac {x^{3}}{24}+\frac {{\mathrm e}^{x}}{5}+\frac {3 \cos \left (2 x \right )}{40}+\frac {3 \sin \left (2 x \right )}{80}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{-4 x} c_{4} +\frac {x^{3}}{24}+\frac {{\mathrm e}^{x}}{5}+\frac {3 \cos \left (2 x \right )}{40}+\frac {3 \sin \left (2 x \right )}{80} \\ \end{align*}
Verification of solutions
\[ y = x^{2} c_{3} +x c_{2} +c_{1} +{\mathrm e}^{-4 x} c_{4} +\frac {x^{3}}{24}+\frac {{\mathrm e}^{x}}{5}+\frac {3 \cos \left (2 x \right )}{40}+\frac {3 \sin \left (2 x \right )}{80} \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -4*_b(_a)+exp(_a)+3*sin(2*_a)+1, _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: 0.016 (sec). Leaf size: 60
dsolve(diff(y(x),x$4)+4*diff(y(x),x$3)=exp(x)+3*sin(2*x)+1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (-\frac {18 \sin \left (x \right )^{2}}{5}+\frac {9 \sin \left (x \right ) \cos \left (x \right )}{5}+x^{3}+\left (12 c_{2} -\frac {18}{5}\right ) x^{2}+\left (24 c_{3} -\frac {9}{5}\right ) x +24 c_{4} \right ) {\mathrm e}^{4 x}+\frac {24 \,{\mathrm e}^{5 x}}{5}-\frac {3 c_{1}}{8}\right ) {\mathrm e}^{-4 x}}{24} \]
✓ Solution by Mathematica
Time used: 0.877 (sec). Leaf size: 59
DSolve[y''''[x]+4*y'''[x]==Exp[x]+3*Sin[2*x]+1,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x^3}{24}+c_4 x^2+\frac {e^x}{5}+\frac {3}{80} \sin (2 x)+\frac {3}{40} \cos (2 x)+c_3 x-\frac {1}{64} c_1 e^{-4 x}+c_2 \]