Internal problem ID [1523]
Internal file name [OUTPUT/1524_Sunday_June_05_2022_02_20_36_AM_21649587/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 26.
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]]
\[ \boxed {2 y^{\prime \prime \prime \prime }-5 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime }-y={\mathrm e}^{x} \left (11+12 x \right )} \] 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 \[ 2 y^{\prime \prime \prime \prime }-5 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime }-y = 0 \] The characteristic equation is \[ 2 \lambda ^{4}-5 \lambda ^{3}+3 \lambda ^{2}+\lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {1}{2}}\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} c_{3} +{\mathrm e}^{-\frac {x}{2}} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{x} x \\ y_3 &= {\mathrm e}^{x} x^{2} \\ y_4 &= {\mathrm e}^{-\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 2 y^{\prime \prime \prime \prime }-5 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime }-y = {\mathrm e}^{x} \left (11+12 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (11+12 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}, {\mathrm e}^{x}, {\mathrm e}^{-\frac {x}{2}}\right \} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}\}] \] Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x^{2}, {\mathrm e}^{x} x^{3}\}] \] Since \({\mathrm e}^{x} x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x^{3}, {\mathrm e}^{x} x^{4}\}] \] 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. \[ y_p = A_{1} {\mathrm e}^{x} x^{3}+A_{2} {\mathrm e}^{x} x^{4} \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ 72 A_{2} {\mathrm e}^{x} x +18 A_{1} {\mathrm e}^{x}+48 A_{2} {\mathrm e}^{x} = {\mathrm e}^{x} \left (11+12 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{6}}, A_{2} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} c_{3} +{\mathrm e}^{-\frac {x}{2}} c_{4}\right ) + \left (\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{6}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-\frac {x}{2}} c_{4} +{\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{6} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\frac {x}{2}} c_{4} +{\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{6} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-\frac {x}{2}} c_{4} +{\mathrm e}^{x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {{\mathrm e}^{x} x^{3}}{6}+\frac {{\mathrm e}^{x} x^{4}}{6} \] 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: 35
dsolve(2*diff(y(x),x$4)-5*diff(y(x),x$3)+3*diff(y(x),x$2)+1*diff(y(x),x)-1*y(x)=exp(x)*(11+12*x),y(x), singsol=all)
\[ y \left (x \right ) = c_{4} {\mathrm e}^{-\frac {x}{2}}+\frac {{\mathrm e}^{x} \left (x^{4}+6 c_{3} x^{2}+x^{3}+6 c_{2} x +6 c_{1} \right )}{6} \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 58
DSolve[2*y''''[x]-5*y'''[x]+3*y''[x]+1*y'[x]-1*y[x]==Exp[x]*(11+12*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (\frac {x^4}{6}+\frac {x^3}{6}+\left (-\frac {1}{3}+c_4\right ) x^2+\left (\frac {4}{9}+c_3\right ) x-\frac {8}{27}+c_2\right )+c_1 e^{-x/2} \]