problem 48

Internal problem ID [13845]
Internal ﬁle name [OUTPUT/13017_Friday_February_23_2024_06_46_37_AM_84591353/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 25. Review exercises for part III. page 447
Problem number: 48.
ODE order: 6.
ODE degree: 1.

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

\[ \boxed {y^{\left (6\right )}-64 y={\mathrm e}^{-2 x}} \] 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 \[ y^{\left (6\right )}-64 y = 0 \] The characteristic equation is \[ \lambda ^{6}-64 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= -2\\ \lambda _3 &= \sqrt {-2-2 i \sqrt {3}}\\ \lambda _4 &= -\sqrt {-2-2 i \sqrt {3}}\\ \lambda _5 &= \sqrt {-2+2 i \sqrt {3}}\\ \lambda _6 &= -\sqrt {-2+2 i \sqrt {3}} \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\sqrt {-2-2 i \sqrt {3}}\, x} c_{3} +{\mathrm e}^{-\sqrt {-2+2 i \sqrt {3}}\, x} c_{4} +{\mathrm e}^{-\sqrt {-2-2 i \sqrt {3}}\, x} c_{5} +{\mathrm e}^{\sqrt {-2+2 i \sqrt {3}}\, x} c_{6} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{2 x} \\ y_3 &= {\mathrm e}^{\sqrt {-2-2 i \sqrt {3}}\, x} \\ y_4 &= {\mathrm e}^{-\sqrt {-2+2 i \sqrt {3}}\, x} \\ y_5 &= {\mathrm e}^{-\sqrt {-2-2 i \sqrt {3}}\, x} \\ y_6 &= {\mathrm e}^{\sqrt {-2+2 i \sqrt {3}}\, x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\left (6\right )}-64 y = {\mathrm e}^{-2 x} \] 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 \[ \left \{{\mathrm e}^{\sqrt {-2-2 i \sqrt {3}}\, x}, {\mathrm e}^{\sqrt {-2+2 i \sqrt {3}}\, x}, {\mathrm e}^{-2 x}, {\mathrm e}^{2 x}, {\mathrm e}^{-\sqrt {-2-2 i \sqrt {3}}\, x}, {\mathrm e}^{-\sqrt {-2+2 i \sqrt {3}}\, x}\right \} \] Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-2 x}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{-2 x} \] 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 \[ -192 A_{1} {\mathrm e}^{-2 x} = {\mathrm e}^{-2 x} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{192}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {x \,{\mathrm e}^{-2 x}}{192} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{\sqrt {-2-2 i \sqrt {3}}\, x} c_{3} +{\mathrm e}^{-\sqrt {-2+2 i \sqrt {3}}\, x} c_{4} +{\mathrm e}^{-\sqrt {-2-2 i \sqrt {3}}\, x} c_{5} +{\mathrm e}^{\sqrt {-2+2 i \sqrt {3}}\, x} c_{6}\right ) + \left (-\frac {x \,{\mathrm e}^{-2 x}}{192}\right ) \\ \end{align*} Which simpliﬁes to \[ y = {\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{-\left (i \sqrt {3}-1\right ) x} c_{3} +{\mathrm e}^{-\left (1+i \sqrt {3}\right ) x} c_{4} +{\mathrm e}^{\left (i \sqrt {3}-1\right ) x} c_{5} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) x} c_{6} -\frac {x \,{\mathrm e}^{-2 x}}{192} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{-\left (i \sqrt {3}-1\right ) x} c_{3} +{\mathrm e}^{-\left (1+i \sqrt {3}\right ) x} c_{4} +{\mathrm e}^{\left (i \sqrt {3}-1\right ) x} c_{5} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) x} c_{6} -\frac {x \,{\mathrm e}^{-2 x}}{192} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{2 x} c_{2} +{\mathrm e}^{-\left (i \sqrt {3}-1\right ) x} c_{3} +{\mathrm e}^{-\left (1+i \sqrt {3}\right ) x} c_{4} +{\mathrm e}^{\left (i \sqrt {3}-1\right ) x} c_{5} +{\mathrm e}^{\left (1+i \sqrt {3}\right ) x} c_{6} -\frac {x \,{\mathrm e}^{-2 x}}{192} \] 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`

\[ y \left (x \right ) = -\frac {\left (\left (-192 c_{3} {\mathrm e}^{3 x}-192 c_{5} {\mathrm e}^{x}\right ) \cos \left (\sqrt {3}\, x \right )+\left (-192 c_{4} {\mathrm e}^{3 x}-192 c_{6} {\mathrm e}^{x}\right ) \sin \left (\sqrt {3}\, x \right )-192 c_{2} {\mathrm e}^{4 x}+x -192 c_{1} \right ) {\mathrm e}^{-2 x}}{192} \]

\[ y(x)\to \frac {1}{768} e^{-2 x} \left (-4 x+768 c_1 e^{4 x}+768 e^x \left (c_2 e^{2 x}+c_3\right ) \cos \left (\sqrt {3} x\right )+768 e^x \left (c_6 e^{2 x}+c_5\right ) \sin \left (\sqrt {3} x\right )-5+768 c_4\right ) \]

17.48 problem 48