Internal problem ID [1553]
Internal file name [OUTPUT/1554_Sunday_June_05_2022_02_22_09_AM_77937519/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 56.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }+4 y=2 \,{\mathrm e}^{x} \left (x +1\right )+{\mathrm e}^{-2 x}} \] 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 \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }+4 y = 0 \] 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 &= -2\\ \lambda _2 &= -2\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+x \,{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+c_{4} {\mathrm e}^{x} x \] 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 &= {\mathrm e}^{x} \\ y_4 &= {\mathrm e}^{x} x \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+2 y^{\prime \prime \prime }-3 y^{\prime \prime }-4 y^{\prime }+4 y = 2 \,{\mathrm e}^{x} \left (x +1\right )+{\mathrm e}^{-2 x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 \,{\mathrm e}^{x} \left (x +1\right )+{\mathrm e}^{-2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{-2 x}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x} x, {\mathrm e}^{-2 x} x, {\mathrm e}^{x}, {\mathrm e}^{-2 x}\} \] Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{-2 x} x\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{-2 x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-2 x}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-2 x}\}, \{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}\}] \] Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-2 x}\}, \{{\mathrm e}^{x} x^{2}, {\mathrm e}^{x} x^{3}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{-2 x}+A_{2} {\mathrm e}^{x} x^{2}+A_{3} {\mathrm e}^{x} x^{3} \] 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 \[ 54 A_{3} {\mathrm e}^{x} x +18 A_{1} {\mathrm e}^{-2 x}+18 A_{2} {\mathrm e}^{x}+36 A_{3} {\mathrm e}^{x} = 2 \,{\mathrm e}^{x} \left (x +1\right )+{\mathrm e}^{-2 x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{18}}, A_{2} = {\frac {1}{27}}, A_{3} = {\frac {1}{27}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{2} {\mathrm e}^{-2 x}}{18}+\frac {{\mathrm e}^{x} x^{2}}{27}+\frac {{\mathrm e}^{x} x^{3}}{27} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+x \,{\mathrm e}^{-2 x} c_{2} +c_{3} {\mathrm e}^{x}+c_{4} {\mathrm e}^{x} x\right ) + \left (\frac {x^{2} {\mathrm e}^{-2 x}}{18}+\frac {{\mathrm e}^{x} x^{2}}{27}+\frac {{\mathrm e}^{x} x^{3}}{27}\right ) \\ \end{align*} Which simplifies to \[ y = \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{3 x}+c_{2} x +c_{1} \right ) {\mathrm e}^{-2 x}+\frac {x^{2} {\mathrm e}^{-2 x}}{18}+\frac {{\mathrm e}^{x} x^{2}}{27}+\frac {{\mathrm e}^{x} x^{3}}{27} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{3 x}+c_{2} x +c_{1} \right ) {\mathrm e}^{-2 x}+\frac {x^{2} {\mathrm e}^{-2 x}}{18}+\frac {{\mathrm e}^{x} x^{2}}{27}+\frac {{\mathrm e}^{x} x^{3}}{27} \\ \end{align*}
Verification of solutions
\[ y = \left (\left (c_{4} x +c_{3} \right ) {\mathrm e}^{3 x}+c_{2} x +c_{1} \right ) {\mathrm e}^{-2 x}+\frac {x^{2} {\mathrm e}^{-2 x}}{18}+\frac {{\mathrm e}^{x} x^{2}}{27}+\frac {{\mathrm e}^{x} x^{3}}{27} \] 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: 50
dsolve(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)=2*exp(x)*(1+x)+exp(-2*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-2 x} \left (\left (x^{3}+x^{2}+\left (27 c_{3} -2\right ) x +27 c_{1} +\frac {10}{9}\right ) {\mathrm e}^{3 x}+\frac {3 x^{2}}{2}+\left (27 c_{4} +2\right ) x +27 c_{2} +1\right )}{27} \]
✓ Solution by Mathematica
Time used: 0.239 (sec). Leaf size: 66
DSolve[y''''[x]+2*y'''[x]-3*y''[x]-4*y'[x]+4*y[x]==2*Exp[x]*(1+x)+Exp[-2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{54} e^{-2 x} \left (3 x^2+(4+54 c_2) x+2+54 c_1\right )+\frac {1}{243} e^x \left (9 x^3+9 x^2+9 (-2+27 c_4) x+10+243 c_3\right ) \]