problem Problem 53

Internal problem ID [12200]
Internal ﬁle name [OUTPUT/10853_Thursday_September_21_2023_05_48_07_AM_23358333/index.tex]

Book: Diﬀerential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS, MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER. Problems page 172
Problem number: Problem 53.
ODE order: 6.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\left (6\right )}-y={\mathrm e}^{2 x}} \] 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 \[ y^{\left (6\right )}-y = 0 \] The characteristic equation is \[ \lambda ^{6}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= \frac {\sqrt {-2-2 i \sqrt {3}}}{2}\\ \lambda _4 &= -\frac {\sqrt {-2-2 i \sqrt {3}}}{2}\\ \lambda _5 &= \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\\ \lambda _6 &= -\frac {\sqrt {-2+2 i \sqrt {3}}}{2} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} c_{3} +{\mathrm e}^{\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} c_{4} +{\mathrm e}^{-\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} c_{5} +{\mathrm e}^{-\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} c_{6} \] 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}^{\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} \\ y_4 &= {\mathrm e}^{\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} \\ y_5 &= {\mathrm e}^{-\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} \\ y_6 &= {\mathrm e}^{-\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (6\right )}-y = {\mathrm e}^{2 x} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{x}, {\mathrm e}^{-x}, {\mathrm e}^{-\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}}, {\mathrm e}^{\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}}, {\mathrm e}^{-\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}}, {\mathrm e}^{\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}}\right \} \] 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. \[ y_p = A_{1} {\mathrm e}^{2 x} \] The unknowns \(\{A_{1}\}\) 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 \[ 63 A_{1} {\mathrm e}^{2 x} = {\mathrm e}^{2 x} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {1}{63}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{2 x}}{63} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} c_{3} +{\mathrm e}^{\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} c_{4} +{\mathrm e}^{-\frac {\sqrt {-2+2 i \sqrt {3}}\, x}{2}} c_{5} +{\mathrm e}^{-\frac {\sqrt {-2-2 i \sqrt {3}}\, x}{2}} c_{6}\right ) + \left (\frac {{\mathrm e}^{2 x}}{63}\right ) \\ \end{align*} Which simpliﬁes to \[ y = {\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{3} +{\mathrm e}^{-\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{4} +{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{5} +{\mathrm e}^{\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{6} +\frac {{\mathrm e}^{2 x}}{63} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{3} +{\mathrm e}^{-\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{4} +{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{5} +{\mathrm e}^{\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{6} +\frac {{\mathrm e}^{2 x}}{63} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{-x} c_{1} +{\mathrm e}^{x} c_{2} +{\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{3} +{\mathrm e}^{-\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{4} +{\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}} c_{5} +{\mathrm e}^{\frac {\left (-1+i \sqrt {3}\right ) x}{2}} c_{6} +\frac {{\mathrm e}^{2 x}}{63} \] 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: 6; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 6; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (x \right ) = {\mathrm e}^{-x} \left (\left (c_{3} {\mathrm e}^{\frac {x}{2}}+c_{5} {\mathrm e}^{\frac {3 x}{2}}\right ) \cos \left (\frac {\sqrt {3}\, x}{2}\right )+\left ({\mathrm e}^{\frac {x}{2}} c_{4} +c_{6} {\mathrm e}^{\frac {3 x}{2}}\right ) \sin \left (\frac {\sqrt {3}\, x}{2}\right )+{\mathrm e}^{2 x} c_{1} +\frac {{\mathrm e}^{3 x}}{63}+c_{2} \right ) \]

\[ y(x)\to \frac {e^{2 x}}{63}+c_1 e^x+c_4 e^{-x}+e^{-x/2} \left (c_2 e^x+c_3\right ) \cos \left (\frac {\sqrt {3} x}{2}\right )+e^{-x/2} \left (c_6 e^x+c_5\right ) \sin \left (\frac {\sqrt {3} x}{2}\right ) \]

2.38 problem Problem 53