Internal problem ID [12773]
Internal file name [OUTPUT/11426_Friday_November_03_2023_06_32_50_AM_61348493/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: 1.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-6 y^{\prime \prime \prime }+13 y^{\prime \prime }-12 y^{\prime }+4 y=2 \,{\mathrm e}^{x}-4 \,{\mathrm e}^{2 x}} \] 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 \[ y^{\prime \prime \prime \prime }-6 y^{\prime \prime \prime }+13 y^{\prime \prime }-12 y^{\prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}-6 \lambda ^{3}+13 \lambda ^{2}-12 \lambda +4 = 0 \] 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 &= x \,{\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-6 y^{\prime \prime \prime }+13 y^{\prime \prime }-12 y^{\prime }+4 y = 2 \,{\mathrm e}^{x}-4 \,{\mathrm e}^{2 x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 \,{\mathrm e}^{x}-4 \,{\mathrm e}^{2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x}\}, \{{\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{x}, x \,{\mathrm e}^{2 x}, {\mathrm e}^{x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}\}, \{{\mathrm e}^{2 x}\}] \] Since \(x \,{\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{x}\}, \{{\mathrm e}^{2 x}\}] \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{x}\}, \{x \,{\mathrm e}^{2 x}\}] \] Since \(x \,{\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{x}\}, \{x^{2} {\mathrm e}^{2 x}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{x}+A_{2} x^{2} {\mathrm e}^{2 x} \] 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 \[ 2 A_{1} {\mathrm e}^{x}+2 A_{2} {\mathrm e}^{2 x} = 2 \,{\mathrm e}^{x}-4 \,{\mathrm e}^{2 x} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 1, A_{2} = -2] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = x^{2} {\mathrm e}^{x}-2 x^{2} {\mathrm e}^{2 x} \] 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 (x^{2} {\mathrm e}^{x}-2 x^{2} {\mathrm e}^{2 x}\right ) \\ \end{align*} Which simplifies to \[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+x^{2} {\mathrm e}^{x}-2 x^{2} {\mathrm e}^{2 x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \right ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+x^{2} {\mathrm e}^{x}-2 x^{2} {\mathrm e}^{2 x} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x +c_{3} \right ) {\mathrm e}^{2 x}+{\mathrm e}^{x} \left (c_{2} x +c_{1} \right )+x^{2} {\mathrm e}^{x}-2 x^{2} {\mathrm e}^{2 x} \] 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] 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: 0.0 (sec). Leaf size: 37
dsolve(diff(y(x),x$4)-6*diff(y(x),x$3)+13*diff(y(x),x$2)-12*diff(y(x),x)+4*y(x)=2*exp(x)-4*exp(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \left (-2 x^{2}+\left (c_{4} +8\right ) x +c_{2} -12\right ) {\mathrm e}^{2 x}+\left (x^{2}+\left (c_{3} +4\right ) x +c_{1} +6\right ) {\mathrm e}^{x} \]
✓ Solution by Mathematica
Time used: 0.187 (sec). Leaf size: 41
DSolve[y''''[x]-6*y'''[x]+13*y''[x]-12*y'[x]+4*y[x]==2*Exp[x]-4*Exp[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (x^2+e^x \left (-2 x^2+(8+c_4) x-12+c_3\right )+(4+c_2) x+6+c_1\right ) \]