Internal problem ID [9861]
Internal file name [OUTPUT/8806_Monday_June_06_2022_05_31_28_AM_66383401/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1537.
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]]
\[ \boxed {y^{\prime \prime \prime \prime }-12 y^{\prime \prime }+12 y=16 \,{\mathrm e}^{x^{2}} x^{4}} \] 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 \[ y^{\prime \prime \prime \prime }-12 y^{\prime \prime }+12 y = 0 \] The characteristic equation is \[ \lambda ^{4}-12 \lambda ^{2}+12 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= \sqrt {6-2 \sqrt {6}}\\ \lambda _2 &= -\sqrt {6-2 \sqrt {6}}\\ \lambda _3 &= \sqrt {6+2 \sqrt {6}}\\ \lambda _4 &= -\sqrt {6+2 \sqrt {6}} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} c_{1} +{\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x} c_{2} +{\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x} c_{3} +{\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} \\ y_2 &= {\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x} \\ y_3 &= {\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x} \\ y_4 &= {\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }-12 y^{\prime \prime }+12 y = 16 \,{\mathrm e}^{x^{2}} x^{4} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 16 \,{\mathrm e}^{x^{2}} x^{4} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x^{2}}, x^{2} {\mathrm e}^{x^{2}}, {\mathrm e}^{x^{2}} x^{3}, {\mathrm e}^{x^{2}} x^{4}, {\mathrm e}^{x^{2}}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x}, {\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x}, {\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x}, {\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x}\right \} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{1} x \,{\mathrm e}^{x^{2}}+A_{2} x^{2} {\mathrm e}^{x^{2}}+A_{3} {\mathrm e}^{x^{2}} x^{3}+A_{4} {\mathrm e}^{x^{2}} x^{4}+A_{5} {\mathrm e}^{x^{2}} \] 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 \[ 192 A_{4} {\mathrm e}^{x^{2}} x^{2}+64 A_{2} x^{4} {\mathrm e}^{x^{2}}+16 A_{4} x^{8} {\mathrm e}^{x^{2}}+24 A_{4} {\mathrm e}^{x^{2}}+48 A_{3} {\mathrm e}^{x^{2}} x +48 A_{2} x^{2} {\mathrm e}^{x^{2}}+144 A_{3} {\mathrm e}^{x^{2}} x^{3}+288 A_{4} {\mathrm e}^{x^{2}} x^{4}+16 A_{3} x^{7} {\mathrm e}^{x^{2}}+16 A_{1} x^{5} {\mathrm e}^{x^{2}}+16 A_{5} x^{4} {\mathrm e}^{x^{2}}+96 A_{3} x^{5} {\mathrm e}^{x^{2}}+128 A_{4} x^{6} {\mathrm e}^{x^{2}}+32 A_{1} x^{3} {\mathrm e}^{x^{2}}+16 A_{2} x^{6} {\mathrm e}^{x^{2}} = 16 \,{\mathrm e}^{x^{2}} x^{4} \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = 0, A_{3} = 0, A_{4} = 0, A_{5} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{x^{2}} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} c_{1} +{\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x} c_{2} +{\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x} c_{3} +{\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x} c_{4}\right ) + \left ({\mathrm e}^{x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} c_{1} +{\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x} c_{2} +{\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x} c_{3} +{\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x} c_{4} +{\mathrm e}^{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} c_{1} +{\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x} c_{2} +{\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x} c_{3} +{\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x} c_{4} +{\mathrm e}^{x^{2}} \] 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: 0.015 (sec). Leaf size: 67
dsolve(diff(diff(diff(diff(y(x),x),x),x),x)-12*diff(diff(y(x),x),x)+12*y(x)-16*x^4*exp(x^2)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{x^{2}}+c_{1} {\mathrm e}^{\sqrt {6-2 \sqrt {6}}\, x}+c_{2} {\mathrm e}^{\sqrt {6+2 \sqrt {6}}\, x}+c_{3} {\mathrm e}^{-\sqrt {6-2 \sqrt {6}}\, x}+c_{4} {\mathrm e}^{-\sqrt {6+2 \sqrt {6}}\, x} \]
✓ Solution by Mathematica
Time used: 1.472 (sec). Leaf size: 93
DSolve[-16*E^x^2*x^4 + 12*y[x] - 12*y''[x] + Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{x^2}+c_1 e^{\sqrt {6-2 \sqrt {6}} x}+c_2 e^{-\sqrt {6-2 \sqrt {6}} x}+c_3 e^{\sqrt {2 \left (3+\sqrt {6}\right )} x}+c_4 e^{-\sqrt {2 \left (3+\sqrt {6}\right )} x} \]