Internal problem ID [1514]
Internal file name [OUTPUT/1515_Sunday_June_05_2022_02_20_17_AM_28180072/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 17.
ODE order: 4.
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
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+3 y^{\prime }-y={\mathrm e}^{x} \left (x^{2}+4 x +3\right )} \] 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 \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+3 y^{\prime }-y = 0 \] The characteristic equation is \[ \lambda ^{4}-2 \lambda ^{3}+3 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =1\right )\\ \lambda _2 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =2\right )\\ \lambda _3 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =3\right )\\ \lambda _4 &= \operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =4\right ) \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =3\right ) x} c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =4\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =1\right ) x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=3\right ) x} \\ y_2 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=4\right ) x} \\ y_3 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=2\right ) x} \\ y_4 &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=1\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+3 y^{\prime }-y = {\mathrm e}^{x} \left (x^{2}+4 x +3\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (x^{2}+4 x +3\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, {\mathrm e}^{x} x^{2}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =1\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =2\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =3\right ) x}, {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =4\right ) x}\} \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} {\mathrm e}^{x} x^{2}+A_{3} {\mathrm e}^{x} \] 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 \[ A_{1} {\mathrm e}^{x}+A_{1} x \,{\mathrm e}^{x}+A_{2} {\mathrm e}^{x} x^{2}+2 A_{2} {\mathrm e}^{x} x +A_{3} {\mathrm e}^{x} = {\mathrm e}^{x} \left (x^{2}+4 x +3\right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 2, A_{2} = 1, A_{3} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = 2 x \,{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=3\right ) x} c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=4\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=1\right ) x} c_{4}\right ) + \left (2 x \,{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}+{\mathrm e}^{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=3\right ) x} c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=4\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} &=1\right ) x} c_{4} +2 x \,{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =3\right ) x} c_{1} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =4\right ) x} c_{2} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =2\right ) x} c_{3} +{\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =1\right ) x} c_{4} +2 x \,{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} \] 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`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 101
dsolve(diff(y(x),x$4)-2*diff(y(x),x$3)+0*diff(y(x),x$2)+3*diff(y(x),x)-y(x)=exp(x)*(3+4*x+x^2),y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =1\right ) x}+c_{2} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =2\right ) x}+c_{3} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =3\right ) x}+c_{4} {\mathrm e}^{\operatorname {RootOf}\left (\textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z} -1, \operatorname {index} =4\right ) x}+\left (x +1\right )^{2} {\mathrm e}^{x} \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 123
DSolve[y''''[x]-2*y'''[x]+0*y''[x]+3*y'[x]-y[x]==Exp[x]*(3+4*x+x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \exp \left (x \text {Root}\left [\text {$\#$1}^4-2 \text {$\#$1}^3+3 \text {$\#$1}-1\&,2\right ]\right )+c_3 \exp \left (x \text {Root}\left [\text {$\#$1}^4-2 \text {$\#$1}^3+3 \text {$\#$1}-1\&,3\right ]\right )+c_4 \exp \left (x \text {Root}\left [\text {$\#$1}^4-2 \text {$\#$1}^3+3 \text {$\#$1}-1\&,4\right ]\right )+c_1 \exp \left (x \text {Root}\left [\text {$\#$1}^4-2 \text {$\#$1}^3+3 \text {$\#$1}-1\&,1\right ]\right )+e^x (x+1)^2 \]