Internal problem ID [1520]
Internal file name [OUTPUT/1521_Sunday_June_05_2022_02_20_29_AM_88806639/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 23.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }-3 y^{\prime \prime }+4 y^{\prime }+4 y={\mathrm e}^{2 x} \left (18 x^{2}+33 x +13\right )} \] 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 \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }-3 y^{\prime \prime }+4 y^{\prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}-2 \lambda ^{3}-3 \lambda ^{2}+4 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -1\\ \lambda _3 &= 2\\ \lambda _4 &= 2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +{\mathrm e}^{2 x} c_{3} +x \,{\mathrm e}^{2 x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= x \,{\mathrm e}^{-x} \\ 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 \[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }-3 y^{\prime \prime }+4 y^{\prime }+4 y = {\mathrm e}^{2 x} \left (18 x^{2}+33 x +13\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{2 x} \left (18 x^{2}+33 x +13\right ) \] 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 \[ \{x \,{\mathrm e}^{-x}, {\mathrm e}^{2 x} x, {\mathrm e}^{-x}, {\mathrm e}^{2 x}\} \] 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 \[ 18 A_{3} {\mathrm e}^{2 x}+36 A_{1} {\mathrm e}^{2 x}+24 A_{2} {\mathrm e}^{2 x}+54 A_{1} x \,{\mathrm e}^{2 x}+108 A_{2} x^{2} {\mathrm e}^{2 x}+144 A_{2} x \,{\mathrm e}^{2 x} = {\mathrm e}^{2 x} \left (18 x^{2}+33 x +13\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = {\frac {1}{6}}, A_{3} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} 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}}{6}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{2 x} \left (c_{4} x +c_{3} \right )+\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{6} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{2 x} \left (c_{4} x +c_{3} \right )+\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{6} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{2} x +c_{1} \right ) {\mathrm e}^{-x}+{\mathrm e}^{2 x} \left (c_{4} x +c_{3} \right )+\frac {x^{3} {\mathrm e}^{2 x}}{6}+\frac {x^{4} {\mathrm e}^{2 x}}{6}+\frac {{\mathrm e}^{2 x} x^{2}}{6} \] 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: 0.016 (sec). Leaf size: 38
dsolve(1*diff(y(x),x$4)-2*diff(y(x),x$3)-3*diff(y(x),x$2)+4*diff(y(x),x)+4*y(x)=exp(2*x)*(13+33*x+18*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{4}+x^{3}+6 c_{4} x +x^{2}+6 c_{2} \right ) {\mathrm e}^{2 x}}{6}+{\mathrm e}^{-x} \left (c_{3} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.104 (sec). Leaf size: 58
DSolve[1*y''''[x]-2*y'''[x]-3*y''[x]+4*y'[x]+4*y[x]==Exp[2*x]*(13+33*x+18*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{54} e^{2 x} \left (9 x^4+9 x^3+9 x^2+18 (-1+3 c_4) x+10+54 c_3\right )+e^{-x} (c_2 x+c_1) \]