Internal problem ID [12776]
Internal file name [OUTPUT/11429_Friday_November_03_2023_06_32_51_AM_75044440/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.4, page 218
Problem number: 4.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime }=6 x -20-120 x^{2} {\mathrm e}^{x}} \] 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 \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime } = 0 \] 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 \[ y_h(x)=c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3} +x^{2} {\mathrm e}^{x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= x \,{\mathrm e}^{x} \\ y_4 &= x^{2} {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-3 y^{\prime \prime \prime }+3 y^{\prime \prime }-y^{\prime } = 6 x -20-120 x^{2} {\mathrm e}^{x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 6 x -20-120 x^{2} {\mathrm e}^{x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, x^{3} {\mathrm e}^{x}\}] \] Since \(x \,{\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}, \{x^{2} {\mathrm e}^{x}, x^{3} {\mathrm e}^{x}, x^{4} {\mathrm e}^{x}\}] \] Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}, \{x^{3} {\mathrm e}^{x}, x^{4} {\mathrm e}^{x}, x^{5} {\mathrm e}^{x}\}] \] 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. \[ y_p = A_{2} x^{2}+A_{1} x +A_{3} x^{3} {\mathrm e}^{x}+A_{4} x^{4} {\mathrm e}^{x}+A_{5} x^{5} {\mathrm e}^{x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}\}\) 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 \[ 6 A_{3} {\mathrm e}^{x}+24 A_{4} {\mathrm e}^{x}+24 A_{4} x \,{\mathrm e}^{x}+60 A_{5} x^{2} {\mathrm e}^{x}+120 A_{5} x \,{\mathrm e}^{x}-A_{1}+6 A_{2}-2 A_{2} x = 6 x -20-120 x^{2} {\mathrm e}^{x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 2, A_{2} = -3, A_{3} = -40, A_{4} = 10, A_{5} = -2] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -3 x^{2}+2 x -40 x^{3} {\mathrm e}^{x}+10 x^{4} {\mathrm e}^{x}-2 x^{5} {\mathrm e}^{x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{x} c_{2} +x \,{\mathrm e}^{x} c_{3} +x^{2} {\mathrm e}^{x} c_{4}\right ) + \left (-3 x^{2}+2 x -40 x^{3} {\mathrm e}^{x}+10 x^{4} {\mathrm e}^{x}-2 x^{5} {\mathrm e}^{x}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} -3 x^{2}+2 x -40 x^{3} {\mathrm e}^{x}+10 x^{4} {\mathrm e}^{x}-2 x^{5} {\mathrm e}^{x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} -3 x^{2}+2 x -40 x^{3} {\mathrm e}^{x}+10 x^{4} {\mathrm e}^{x}-2 x^{5} {\mathrm e}^{x} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x^{2}+c_{3} x +c_{2} \right ) {\mathrm e}^{x}+c_{1} -3 x^{2}+2 x -40 x^{3} {\mathrm e}^{x}+10 x^{4} {\mathrm e}^{x}-2 x^{5} {\mathrm e}^{x} \] 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] -> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = -120*_a^2*exp(_a)-3*(diff(_b(_a), _a))+3*(diff(diff(_b(_a), 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: 0.0 (sec). Leaf size: 56
dsolve(diff(y(x),x$4)-3*diff(y(x),x$3)+3*diff(y(x),x$2)-diff(y(x),x)=6*x-20-120*x^2*exp(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (-2 x^{5}+10 x^{4}-40 x^{3}+\left (c_{3} +120\right ) x^{2}+\left (c_{2} -2 c_{3} -240\right ) x +c_{1} -c_{2} +2 c_{3} +240\right ) {\mathrm e}^{x}-3 x^{2}+2 x +c_{4} \]
✓ Solution by Mathematica
Time used: 0.569 (sec). Leaf size: 65
DSolve[y''''[x]-3*y'''[x]+3*y''[x]-y'[x]==6*x-20-120*x^2*Exp[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -3 x^2+e^x \left (-2 x^5+10 x^4-40 x^3+(120+c_3) x^2+(-240+c_2-2 c_3) x+240+c_1-c_2+2 c_3\right )+2 x+c_4 \]