Internal problem ID [1529]
Internal file name [OUTPUT/1530_Sunday_June_05_2022_02_20_50_AM_91271860/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 32.
ODE order: 3.
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
[[_3rd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime }-2 y=-{\mathrm e}^{x} \left (\left (4 x^{2}+5 x +9\right ) \cos \left (2 x \right )-\left (-3 x^{2}-5 x +6\right ) \sin \left (2 x \right )\right )} \] 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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime }-2 y = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}+\lambda -2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= i\\ \lambda _3 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{2 x} c_{1} +{\mathrm e}^{i x} c_{2} +{\mathrm e}^{-i x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{2 x} \\ y_2 &= {\mathrm e}^{i x} \\ y_3 &= {\mathrm e}^{-i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }+y^{\prime }-2 y = -{\mathrm e}^{x} \left (\left (4 x^{2}+5 x +9\right ) \cos \left (2 x \right )-\left (-3 x^{2}-5 x +6\right ) \sin \left (2 x \right )\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ -{\mathrm e}^{x} \left (\left (4 x^{2}+5 x +9\right ) \cos \left (2 x \right )-\left (-3 x^{2}-5 x +6\right ) \sin \left (2 x \right )\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x} \cos \left (2 x \right ), {\mathrm e}^{x} \sin \left (2 x \right ), {\mathrm e}^{x} \cos \left (2 x \right ) x, {\mathrm e}^{x} \cos \left (2 x \right ) x^{2}, {\mathrm e}^{x} \sin \left (2 x \right ) x, {\mathrm e}^{x} \sin \left (2 x \right ) x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{i x}, {\mathrm e}^{2 x}, {\mathrm e}^{-i x}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{x} \cos \left (2 x \right )+A_{2} {\mathrm e}^{x} \sin \left (2 x \right )+A_{3} {\mathrm e}^{x} \cos \left (2 x \right ) x +A_{4} {\mathrm e}^{x} \cos \left (2 x \right ) x^{2}+A_{5} {\mathrm e}^{x} \sin \left (2 x \right ) x +A_{6} {\mathrm e}^{x} \sin \left (2 x \right ) x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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 \[ -24 A_{4} {\mathrm e}^{x} \cos \left (2 x \right ) x -24 A_{6} {\mathrm e}^{x} \sin \left (2 x \right ) x -8 A_{4} {\mathrm e}^{x} \sin \left (2 x \right ) x +2 A_{6} {\mathrm e}^{x} \sin \left (2 x \right )-12 A_{4} {\mathrm e}^{x} \sin \left (2 x \right )-4 A_{3} {\mathrm e}^{x} \sin \left (2 x \right )+2 A_{4} {\mathrm e}^{x} \cos \left (2 x \right )+4 A_{5} {\mathrm e}^{x} \cos \left (2 x \right )-12 A_{3} {\mathrm e}^{x} \cos \left (2 x \right )+8 A_{3} {\mathrm e}^{x} \sin \left (2 x \right ) x +8 A_{4} {\mathrm e}^{x} \sin \left (2 x \right ) x^{2}-12 A_{5} {\mathrm e}^{x} \sin \left (2 x \right )-8 A_{5} {\mathrm e}^{x} \cos \left (2 x \right ) x -8 A_{6} {\mathrm e}^{x} \cos \left (2 x \right ) x^{2}-6 A_{4} {\mathrm e}^{x} \cos \left (2 x \right ) x^{2}-6 A_{5} {\mathrm e}^{x} \sin \left (2 x \right ) x -6 A_{6} {\mathrm e}^{x} \sin \left (2 x \right ) x^{2}-6 A_{3} {\mathrm e}^{x} \cos \left (2 x \right ) x +8 A_{6} {\mathrm e}^{x} \cos \left (2 x \right ) x -6 A_{2} {\mathrm e}^{x} \sin \left (2 x \right )-6 A_{1} {\mathrm e}^{x} \cos \left (2 x \right )+8 A_{1} {\mathrm e}^{x} \sin \left (2 x \right )-8 A_{2} {\mathrm e}^{x} \cos \left (2 x \right )+12 A_{6} {\mathrm e}^{x} \cos \left (2 x \right ) = -{\mathrm e}^{x} \left (\left (4 x^{2}+5 x +9\right ) \cos \left (2 x \right )-\left (-3 x^{2}-5 x +6\right ) \sin \left (2 x \right )\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {61}{50}}, A_{2} = -{\frac {27}{50}}, A_{3} = {\frac {11}{10}}, A_{4} = 0, A_{5} = {\frac {3}{10}}, A_{6} = {\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {61 \,{\mathrm e}^{x} \cos \left (2 x \right )}{50}-\frac {27 \,{\mathrm e}^{x} \sin \left (2 x \right )}{50}+\frac {11 \,{\mathrm e}^{x} \cos \left (2 x \right ) x}{10}+\frac {3 \,{\mathrm e}^{x} \sin \left (2 x \right ) x}{10}+\frac {{\mathrm e}^{x} \sin \left (2 x \right ) x^{2}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{2 x} c_{1} +{\mathrm e}^{i x} c_{2} +{\mathrm e}^{-i x} c_{3}\right ) + \left (\frac {61 \,{\mathrm e}^{x} \cos \left (2 x \right )}{50}-\frac {27 \,{\mathrm e}^{x} \sin \left (2 x \right )}{50}+\frac {11 \,{\mathrm e}^{x} \cos \left (2 x \right ) x}{10}+\frac {3 \,{\mathrm e}^{x} \sin \left (2 x \right ) x}{10}+\frac {{\mathrm e}^{x} \sin \left (2 x \right ) x^{2}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{2 x} c_{1} +{\mathrm e}^{i x} c_{2} +{\mathrm e}^{-i x} c_{3} +\frac {61 \,{\mathrm e}^{x} \cos \left (2 x \right )}{50}-\frac {27 \,{\mathrm e}^{x} \sin \left (2 x \right )}{50}+\frac {11 \,{\mathrm e}^{x} \cos \left (2 x \right ) x}{10}+\frac {3 \,{\mathrm e}^{x} \sin \left (2 x \right ) x}{10}+\frac {{\mathrm e}^{x} \sin \left (2 x \right ) x^{2}}{2} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{2 x} c_{1} +{\mathrm e}^{i x} c_{2} +{\mathrm e}^{-i x} c_{3} +\frac {61 \,{\mathrm e}^{x} \cos \left (2 x \right )}{50}-\frac {27 \,{\mathrm e}^{x} \sin \left (2 x \right )}{50}+\frac {11 \,{\mathrm e}^{x} \cos \left (2 x \right ) x}{10}+\frac {3 \,{\mathrm e}^{x} \sin \left (2 x \right ) x}{10}+\frac {{\mathrm e}^{x} \sin \left (2 x \right ) x^{2}}{2} \] Verified OK.
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 48
dsolve(1*diff(y(x),x$3)-2*diff(y(x),x$2)+1*diff(y(x),x)-2*y(x)=-exp(x)*((9+5*x+4*x^2)*cos(2*x)-(6-5*x-3*x^2)*sin(2*x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{x} \left (x^{2}+\frac {3}{5} x -\frac {27}{25}\right ) \sin \left (2 x \right )}{2}+\frac {\left (55 x +61\right ) {\mathrm e}^{x} \cos \left (2 x \right )}{50}+\cos \left (x \right ) c_{1} +\sin \left (x \right ) c_{2} +c_{3} {\mathrm e}^{2 x} \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 65
DSolve[1*y'''[x]-2*y''[x]+1*y'[x]-2*y[x]==Exp[2*x]*((9+5*x+4*x^2)*Cos[2*x]-(6-5*x-3*x^2)*Sin[2*x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {e^{2 x} \left (\left (6760 x^2-17680 x-29907\right ) \sin (2 x)+2 \left (5915 x^2+7345 x+3928\right ) \cos (2 x)\right )}{43940}+c_3 e^{2 x}+c_1 \cos (x)+c_2 \sin (x) \]