problem section 9.3, problem 24

Internal problem ID [1521]
Internal ﬁle name [OUTPUT/1522_Sunday_June_05_2022_02_20_32_AM_57783203/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 24.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+4 y^{\prime }={\mathrm e}^{2 x} \left (12 x^{2}+26 x +15\right )} \] 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 \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+4 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{4}-3 \lambda ^{3}+4 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -1\\ \lambda _3 &= 2\\ \lambda _4 &= 2 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +c_{2} +{\mathrm e}^{2 x} c_{3} +x \,{\mathrm e}^{2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= {\mathrm e}^{2 x} \\ y_4 &= {\mathrm e}^{2 x} x \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+4 y^{\prime } = {\mathrm e}^{2 x} \left (12 x^{2}+26 x +15\right ) \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \left (12 x^{2}+26 x +15\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x} x^{2}, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{2 x} x, {\mathrm e}^{-x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x, {\mathrm e}^{2 x} x^{2}\}] \] Since \({\mathrm e}^{2 x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{2 x}, x^{4} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x^{2}\}] \] 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. \[ y_p = A_{1} x^{3} {\mathrm e}^{2 x}+A_{2} x^{4} {\mathrm e}^{2 x}+A_{3} {\mathrm e}^{2 x} x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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 \[ 12 A_{3} {\mathrm e}^{2 x}+30 A_{1} {\mathrm e}^{2 x}+24 A_{2} {\mathrm e}^{2 x}+36 A_{1} x \,{\mathrm e}^{2 x}+72 A_{2} x^{2} {\mathrm e}^{2 x}+120 A_{2} x \,{\mathrm e}^{2 x} = {\mathrm e}^{2 x} \left (12 x^{2}+26 x +15\right ) \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = {\frac {1}{6}}, A_{3} = {\frac {1}{2}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +c_{2} +{\mathrm e}^{2 x} c_{3} +x \,{\mathrm e}^{2 x} c_{4}\right ) + \left (\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{2}\right ) \\ \end{align*} Which simpliﬁes to \[ y = {\mathrm e}^{2 x} \left (c_{4} x +c_{3} \right )+{\mathrm e}^{-x} c_{1} +c_{2} +\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{2} \]

Summary

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

Veriﬁcation of solutions

\[ y = {\mathrm e}^{2 x} \left (c_{4} x +c_{3} \right )+{\mathrm e}^{-x} c_{1} +c_{2} +\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{2} \] 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] 
-> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = 12*exp(2*_a)*_a^2+26*exp(2*_a)*_a+15*exp(2*_a)-4*_b(_a)+3*( 
   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 
<- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`

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

\[ y(x)\to \frac {1}{12} e^{2 x} \left (2 x^4+2 x^3+6 x^2+6 (-2+c_3) x+8+6 c_2-3 c_3\right )+c_1 \left (-e^{-x}\right )+c_4 \]

19.24 problem section 9.3, problem 24