Internal problem ID [1524]
Internal file name [OUTPUT/1525_Sunday_June_05_2022_02_20_38_AM_56632814/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 27.
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, _missing_y]]
\[ \boxed {y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime }={\mathrm e}^{-x} \left (10 x^{2}-24 x +5\right )} \] 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 \[ y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+3 \lambda ^{3}+3 \lambda ^{2}+\lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -1\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3} +c_{4} \] 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}^{-x} x^{2} \\ y_4 &= 1 \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }+3 y^{\prime \prime }+y^{\prime } = {\mathrm e}^{-x} \left (10 x^{2}-24 x +5\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{-x} \left (10 x^{2}-24 x +5\right ) \] 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 \[ \{1, x \,{\mathrm e}^{-x}, {\mathrm e}^{-x} x^{2}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}, {\mathrm e}^{-x} x^{2}\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{-x}, x^{4} {\mathrm e}^{-x}, {\mathrm e}^{-x} x^{2}\}] \] Since \({\mathrm e}^{-x} x^{2}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{-x}, x^{4} {\mathrm e}^{-x}, x^{5} {\mathrm e}^{-x}\}] \] 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. \[ y_p = A_{1} x^{3} {\mathrm e}^{-x}+A_{2} x^{4} {\mathrm e}^{-x}+A_{3} x^{5} {\mathrm e}^{-x} \] 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 \[ 24 A_{2} {\mathrm e}^{-x}-6 A_{1} {\mathrm e}^{-x}-24 A_{2} x \,{\mathrm e}^{-x}-60 A_{3} x^{2} {\mathrm e}^{-x}+120 A_{3} x \,{\mathrm e}^{-x} = {\mathrm e}^{-x} \left (10 x^{2}-24 x +5\right ) \] 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 \[ y_p = -\frac {x^{3} {\mathrm e}^{-x}}{6}+\frac {x^{4} {\mathrm e}^{-x}}{6}-\frac {x^{5} {\mathrm e}^{-x}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3} +c_{4}\right ) + \left (-\frac {x^{3} {\mathrm e}^{-x}}{6}+\frac {x^{4} {\mathrm e}^{-x}}{6}-\frac {x^{5} {\mathrm e}^{-x}}{6}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+c_{4} -\frac {x^{3} {\mathrm e}^{-x}}{6}+\frac {x^{4} {\mathrm e}^{-x}}{6}-\frac {x^{5} {\mathrm e}^{-x}}{6} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+c_{4} -\frac {x^{3} {\mathrm e}^{-x}}{6}+\frac {x^{4} {\mathrm e}^{-x}}{6}-\frac {x^{5} {\mathrm e}^{-x}}{6} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+c_{4} -\frac {x^{3} {\mathrm e}^{-x}}{6}+\frac {x^{4} {\mathrm e}^{-x}}{6}-\frac {x^{5} {\mathrm e}^{-x}}{6} \] 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] -> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = 10*exp(-_a)*_a^2-24*_a*exp(-_a)+5*exp(-_a)-_b(_a)-3*(diff(_ Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful <- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 55
dsolve(1*diff(y(x),x$4)+3*diff(y(x),x$3)+3*diff(y(x),x$2)+1*diff(y(x),x)-0*y(x)=exp(-x)*(5-24*x+10*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-x^{5}+x^{4}-x^{3}+\left (-6 c_{3} -3\right ) x^{2}+\left (-6 c_{2} -12 c_{3} -6\right ) x -6 c_{1} -6 c_{2} -12 c_{3} -6\right ) {\mathrm e}^{-x}}{6}+c_{4} \]
✓ Solution by Mathematica
Time used: 0.036 (sec). Leaf size: 65
DSolve[1*y''''[x]+3*y'''[x]+3*y''[x]+1*y'[x]-0*y[x]==Exp[-x]*(5-24*x+10*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{6} e^{-x} \left (-x^5+x^4-x^3-3 (1+2 c_3) x^2-6 (1+c_2+2 c_3) x-6 (1+c_1+c_2+2 c_3)\right )+c_4 \]