problem 16.1 (iii)

Internal problem ID [12045]
Internal ﬁle name [OUTPUT/10698_Sunday_September_03_2023_12_36_41_PM_90112867/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 16, Higher order linear equations with constant coeﬃcients. Exercises page 153
Problem number: 16.1 (iii).
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {x^{\prime \prime \prime \prime }-4 x^{\prime \prime \prime }+8 x^{\prime \prime }-8 x^{\prime }+4 x=\sin \left (t \right )} \] This is higher order nonhomogeneous ODE. Let the solution be \[ x = x_h + x_p \] Where \(x_h\) is the solution to the homogeneous ODE And \(x_p\) is a particular solution to the nonhomogeneous ODE. \(x_h\) is the solution to \[ x^{\prime \prime \prime \prime }-4 x^{\prime \prime \prime }+8 x^{\prime \prime }-8 x^{\prime }+4 x = 0 \] The characteristic equation is \[ \lambda ^{4}-4 \lambda ^{3}+8 \lambda ^{2}-8 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1-i\\ \lambda _2 &= 1+i\\ \lambda _3 &= 1-i\\ \lambda _4 &= 1+i \end {align*}

Therefore the homogeneous solution is \[ x_h(t)={\mathrm e}^{\left (1+i\right ) t} c_{1} +t \,{\mathrm e}^{\left (1+i\right ) t} c_{2} +{\mathrm e}^{\left (1-i\right ) t} c_{3} +t \,{\mathrm e}^{\left (1-i\right ) t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} x_1 &= {\mathrm e}^{\left (1+i\right ) t} \\ x_2 &= t \,{\mathrm e}^{\left (1+i\right ) t} \\ x_3 &= {\mathrm e}^{\left (1-i\right ) t} \\ x_4 &= t \,{\mathrm e}^{\left (1-i\right ) t} \\ \end{align*} Now the particular solution to the given ODE is found \[ x^{\prime \prime \prime \prime }-4 x^{\prime \prime \prime }+8 x^{\prime \prime }-8 x^{\prime }+4 x = \sin \left (t \right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ \sin \left (t \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{\cos \left (t \right ), \sin \left (t \right )\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{t \,{\mathrm e}^{\left (1-i\right ) t}, t \,{\mathrm e}^{\left (1+i\right ) t}, {\mathrm e}^{\left (1-i\right ) t}, {\mathrm e}^{\left (1+i\right ) t}\} \] 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. \[ x_p = A_{1} \cos \left (t \right )+A_{2} \sin \left (t \right ) \] The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(x_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ -3 A_{1} \cos \left (t \right )-3 A_{2} \sin \left (t \right )+4 A_{1} \sin \left (t \right )-4 A_{2} \cos \left (t \right ) = \sin \left (t \right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {4}{25}}, A_{2} = -{\frac {3}{25}}\right ] \] Substituting the above back in the above trial solution \(x_p\), gives the particular solution \[ x_p = \frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25} \] Therefore the general solution is \begin{align*} x &= x_h + x_p \\ &= \left ({\mathrm e}^{\left (1+i\right ) t} c_{1} +t \,{\mathrm e}^{\left (1+i\right ) t} c_{2} +{\mathrm e}^{\left (1-i\right ) t} c_{3} +t \,{\mathrm e}^{\left (1-i\right ) t} c_{4}\right ) + \left (\frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25}\right ) \\ \end{align*} Which simpliﬁes to \[ x = \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (1-i\right ) t}+{\mathrm e}^{\left (1+i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (1-i\right ) t}+{\mathrm e}^{\left (1+i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25} \\ \end{align*}

Veriﬁcation of solutions

\[ x = \left (c_{4} t +c_{3} \right ) {\mathrm e}^{\left (1-i\right ) t}+{\mathrm e}^{\left (1+i\right ) t} \left (c_{2} t +c_{1} \right )+\frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25} \] 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`

\[ x \left (t \right ) = \left (\left (c_{3} t +c_{1} \right ) \cos \left (t \right )+\sin \left (t \right ) \left (c_{4} t +c_{2} \right )\right ) {\mathrm e}^{t}+\frac {4 \cos \left (t \right )}{25}-\frac {3 \sin \left (t \right )}{25} \]

\[ x(t)\to \left (\frac {4}{25}+e^t (c_4 t+c_3)\right ) \cos (t)+\left (-\frac {3}{25}+e^t (c_2 t+c_1)\right ) \sin (t) \]

9.3 problem 16.1 (iii)