problem 47

Internal problem ID [11820]
Internal ﬁle name [OUTPUT/11830_Thursday_April_11_2024_08_51_26_PM_89373344/index.tex]

Book: Diﬀerential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section: Chapter 4, Section 4.3. The method of undetermined coeﬃcients. Exercises page 151
Problem number: 47.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime }-6 y^{\prime \prime }+12 y^{\prime }-8 y={\mathrm e}^{2 x} x +{\mathrm e}^{3 x} x^{2}} \] 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 \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+12 y^{\prime }-8 y = 0 \] The characteristic equation is \[ \lambda ^{3}-6 \lambda ^{2}+12 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 2\\ \lambda _3 &= 2 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{2 x}+c_{2} {\mathrm e}^{2 x} x +x^{2} {\mathrm e}^{2 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{2 x} \\ y_2 &= {\mathrm e}^{2 x} x \\ y_3 &= x^{2} {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+12 y^{\prime }-8 y = {\mathrm e}^{2 x} x +{\mathrm e}^{3 x} x^{2} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} x +{\mathrm e}^{3 x} x^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}, \{x \,{\mathrm e}^{3 x}, {\mathrm e}^{3 x} x^{2}, {\mathrm e}^{3 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x\}, \{x \,{\mathrm e}^{3 x}, {\mathrm e}^{3 x} x^{2}, {\mathrm e}^{3 x}\}] \] Since \({\mathrm e}^{2 x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{2 x}, x^{3} {\mathrm e}^{2 x}\}, \{x \,{\mathrm e}^{3 x}, {\mathrm e}^{3 x} x^{2}, {\mathrm e}^{3 x}\}] \] Since \(x^{2} {\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{4}\}, \{x \,{\mathrm e}^{3 x}, {\mathrm e}^{3 x} x^{2}, {\mathrm e}^{3 x}\}] \] 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. \[ y_p = A_{1} x^{3} {\mathrm e}^{2 x}+A_{2} {\mathrm e}^{2 x} x^{4}+A_{3} x \,{\mathrm e}^{3 x}+A_{4} {\mathrm e}^{3 x} x^{2}+A_{5} {\mathrm e}^{3 x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) 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 \[ 24 A_{2} {\mathrm e}^{2 x} x +6 A_{4} {\mathrm e}^{3 x} x +6 A_{4} {\mathrm e}^{3 x}+6 A_{1} {\mathrm e}^{2 x}+A_{3} x \,{\mathrm e}^{3 x}+A_{4} {\mathrm e}^{3 x} x^{2}+A_{5} {\mathrm e}^{3 x}+3 A_{3} {\mathrm e}^{3 x} = {\mathrm e}^{2 x} x +{\mathrm e}^{3 x} x^{2} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{24}}, A_{3} = -6, A_{4} = 1, A_{5} = 12\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{2 x} x^{4}}{24}-6 x \,{\mathrm e}^{3 x}+{\mathrm e}^{3 x} x^{2}+12 \,{\mathrm e}^{3 x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{2 x}+c_{2} {\mathrm e}^{2 x} x +x^{2} {\mathrm e}^{2 x} c_{3}\right ) + \left (\frac {{\mathrm e}^{2 x} x^{4}}{24}-6 x \,{\mathrm e}^{3 x}+{\mathrm e}^{3 x} x^{2}+12 \,{\mathrm e}^{3 x}\right ) \\ \end{align*} Which simpliﬁes to \[ y = {\mathrm e}^{2 x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{2 x} x^{4}}{24}-6 x \,{\mathrm e}^{3 x}+{\mathrm e}^{3 x} x^{2}+12 \,{\mathrm e}^{3 x} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{2 x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{2 x} x^{4}}{24}-6 x \,{\mathrm e}^{3 x}+{\mathrm e}^{3 x} x^{2}+12 \,{\mathrm e}^{3 x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{2 x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{2 x} x^{4}}{24}-6 x \,{\mathrm e}^{3 x}+{\mathrm e}^{3 x} x^{2}+12 \,{\mathrm e}^{3 x} \] 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`

\[ y \left (x \right ) = \frac {\left (x^{4}+24 c_{3} x^{2}+24 c_{2} x +24 c_{1} \right ) {\mathrm e}^{2 x}}{24}+{\mathrm e}^{3 x} \left (x^{2}-6 x +12\right ) \]

\[ y(x)\to \frac {1}{24} e^{2 x} \left (x^4+24 e^x \left (x^2-6 x+12\right )+24 c_3 x^2+24 c_2 x+24 c_1\right ) \]

11.47 problem 47