problem 42

Internal problem ID [7483]
Internal ﬁle name [OUTPUT/6450_Sunday_June_19_2022_05_03_01_PM_44498961/index.tex]

Book: Second order enumerated odes
Section: section 2
Problem number: 42.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y=x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x}} \] 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 \[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y = 0 \] The characteristic equation is \[ \lambda ^{4}-\lambda ^{3}-3 \lambda ^{2}+5 \lambda -2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -2\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3} +x^{2} {\mathrm e}^{x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= x \,{\mathrm e}^{x} \\ y_4 &= x^{2} {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y = x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}, {\mathrm e}^{-2 x}\} \] Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] Since \(x \,{\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{x^{2} {\mathrm e}^{x}, x^{3} {\mathrm e}^{x}\}] \] Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{x^{3} {\mathrm e}^{x}, {\mathrm e}^{x} x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{-2 x} x +A_{2} x^{3} {\mathrm e}^{x}+A_{3} {\mathrm e}^{x} x^{4} \] 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 \[ 72 A_{3} {\mathrm e}^{x} x -27 A_{1} {\mathrm e}^{-2 x}+18 A_{2} {\mathrm e}^{x}+24 A_{3} {\mathrm e}^{x} = x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -{\frac {1}{9}}, A_{2} = -{\frac {1}{54}}, A_{3} = {\frac {1}{72}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {{\mathrm e}^{-2 x} x}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {{\mathrm e}^{x} x^{4}}{72} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 x} c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3} +x^{2} {\mathrm e}^{x} c_{4}\right ) + \left (-\frac {{\mathrm e}^{-2 x} x}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {{\mathrm e}^{x} x^{4}}{72}\right ) \\ \end{align*} Which simpliﬁes to \[ y = \left (\left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x} x}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {{\mathrm e}^{x} x^{4}}{72} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x} x}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {{\mathrm e}^{x} x^{4}}{72} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (\left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{3 x}+c_{1} \right ) {\mathrm e}^{-2 x}-\frac {{\mathrm e}^{-2 x} x}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {{\mathrm e}^{x} x^{4}}{72} \] 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`

\[ y \left (x \right ) = \frac {\left (\left (x^{4}-\frac {4 x^{3}}{3}+\left (72 c_{4} +\frac {4}{3}\right ) x^{2}+\left (72 c_{3} -\frac {8}{9}\right ) x +72 c_{1} +\frac {8}{27}\right ) {\mathrm e}^{3 x}-8 x +72 c_{2} -8\right ) {\mathrm e}^{-2 x}}{72} \]

\[ y(x)\to e^x \left (\frac {x^4}{72}-\frac {x^3}{54}+\left (\frac {1}{54}+c_4\right ) x^2+\left (-\frac {1}{81}+c_3\right ) x+\frac {1}{243}+c_2\right )-\frac {1}{9} e^{-2 x} (x+1-9 c_1) \]

2.42 problem 42