Internal problem ID [5391]
Internal file name [OUTPUT/4882_Sunday_February_04_2024_12_46_46_AM_39250907/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 15. Linear equations with constant coefficients (Variation of parameters).
Supplemetary problems. Page 98
Problem number: 21.
ODE order: 3.
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
[[_3rd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime }-4 y^{\prime }+4 y=2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x}} \] 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 \[ y^{\prime \prime \prime }-y^{\prime \prime }-4 y^{\prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{3}-\lambda ^{2}-4 \lambda +4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 2\\ \lambda _3 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-y^{\prime \prime }-4 y^{\prime }+4 y = 2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}\}, \{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-2 x}, {\mathrm e}^{2 x}\} \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{1, x, x^{2}\}, \{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x, {\mathrm e}^{2 x} x^{3}\}] \] 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. \[ y_p = A_{3} x^{2}+A_{2} x +A_{1}+A_{4} x^{2} {\mathrm e}^{2 x}+A_{5} {\mathrm e}^{2 x} x +A_{6} {\mathrm e}^{2 x} x^{3} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\}\) 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 \[ 12 A_{6} {\mathrm e}^{2 x} x^{2}+30 A_{6} {\mathrm e}^{2 x} x +8 A_{4} x \,{\mathrm e}^{2 x}-8 A_{3} x -2 A_{3}+4 A_{1}-4 A_{2}+10 A_{4} {\mathrm e}^{2 x}+6 A_{6} {\mathrm e}^{2 x}+4 A_{2} x +4 A_{3} x^{2}+4 A_{5} {\mathrm e}^{2 x} = 2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = 0, A_{2} = 0, A_{3} = {\frac {1}{2}}, A_{4} = 0, A_{5} = 0, A_{6} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{2}}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3}\right ) + \left (\frac {x^{2}}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{6}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} +\frac {x^{2}}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} +\frac {x^{2}}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{6} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-y^{\prime \prime }-4 y^{\prime }+4 y=2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +2 x^{2}+y_{3}\left (x \right )+4 y_{2}\left (x \right )-4 y_{1}\left (x \right )+{\mathrm e}^{2 x}-4 x -1 \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +2 x^{2}+y_{3}\left (x \right )+4 y_{2}\left (x \right )-4 y_{1}\left (x \right )+{\mathrm e}^{2 x}-4 x -1\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -4 & 4 & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 2 x^{2}-4 x -1+2 x^{2} {\mathrm e}^{2 x}+5 \,{\mathrm e}^{2 x} x +{\mathrm e}^{2 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -4 & 4 & 1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{x} & {\mathrm e}^{2 x} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{x} & {\mathrm e}^{2 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & 1 & \frac {1}{4} \\ -\frac {1}{2} & 1 & \frac {1}{2} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} -\frac {\left (3 \,{\mathrm e}^{4 x}-8 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{6} & -\frac {{\mathrm e}^{-2 x}}{4}+\frac {{\mathrm e}^{2 x}}{4} & \frac {\left (3 \,{\mathrm e}^{4 x}-4 \,{\mathrm e}^{3 x}+1\right ) {\mathrm e}^{-2 x}}{12} \\ -\frac {\left (3 \,{\mathrm e}^{4 x}-4 \,{\mathrm e}^{3 x}+1\right ) {\mathrm e}^{-2 x}}{3} & \frac {{\mathrm e}^{-2 x}}{2}+\frac {{\mathrm e}^{2 x}}{2} & \frac {\left (3 \,{\mathrm e}^{4 x}-2 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{6} \\ -\frac {2 \left (3 \,{\mathrm e}^{4 x}-2 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{3} & -{\mathrm e}^{-2 x}+{\mathrm e}^{2 x} & -\frac {\left (-3 \,{\mathrm e}^{4 x}+{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{3} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {\left (2 x^{3} {\mathrm e}^{4 x}-3 \,{\mathrm e}^{4 x}+6 x^{2} {\mathrm e}^{2 x}+4 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{12} \\ \frac {\left (2 x^{3} {\mathrm e}^{4 x}+3 \,{\mathrm e}^{4 x} x^{2}-3 \,{\mathrm e}^{4 x}+2 \,{\mathrm e}^{3 x}+6 \,{\mathrm e}^{2 x} x +1\right ) {\mathrm e}^{-2 x}}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (2 x^{3}+6 x^{2}+3 x -3\right ) {\mathrm e}^{4 x}+3 \,{\mathrm e}^{2 x}+{\mathrm e}^{3 x}-1\right )}{3} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} \frac {\left (2 x^{3} {\mathrm e}^{4 x}-3 \,{\mathrm e}^{4 x}+6 x^{2} {\mathrm e}^{2 x}+4 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{12} \\ \frac {\left (2 x^{3} {\mathrm e}^{4 x}+3 \,{\mathrm e}^{4 x} x^{2}-3 \,{\mathrm e}^{4 x}+2 \,{\mathrm e}^{3 x}+6 \,{\mathrm e}^{2 x} x +1\right ) {\mathrm e}^{-2 x}}{6} \\ \frac {{\mathrm e}^{-2 x} \left (\left (2 x^{3}+6 x^{2}+3 x -3\right ) {\mathrm e}^{4 x}+3 \,{\mathrm e}^{2 x}+{\mathrm e}^{3 x}-1\right )}{3} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {{\mathrm e}^{-2 x} \left (\left (2 x^{3}+3 c_{3} -3\right ) {\mathrm e}^{4 x}+6 x^{2} {\mathrm e}^{2 x}+12 c_{2} {\mathrm e}^{3 x}+3 c_{1} +4 \,{\mathrm e}^{3 x}-1\right )}{12} \end {array} \]
Maple trace
`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`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 42
dsolve(diff(y(x),x$3)-diff(y(x),x$2)-4*diff(y(x),x)+4*y(x)=2*x^2-4*x-1+2*x^2*exp(2*x)+5*x*exp(2*x)+exp(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (x^{3}+6 c_{3} \right ) {\mathrm e}^{4 x}+3 \,{\mathrm e}^{2 x} x^{2}+6 \,{\mathrm e}^{3 x} c_{1} +6 c_{2} \right ) {\mathrm e}^{-2 x}}{6} \]
✓ Solution by Mathematica
Time used: 0.519 (sec). Leaf size: 44
DSolve[y'''[x]-y''[x]-4*y'[x]+4*y[x]==2*x^2-4*x-1+2*x^2*Exp[2*x]+5*x*Exp[2*x]+Exp[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{6} \left (e^{2 x} x+3\right ) x^2+c_1 e^{-2 x}+c_2 e^x+c_3 e^{2 x} \]