Internal problem ID [9135]
Internal file name [OUTPUT/8070_Monday_June_06_2022_01_39_17_AM_63457504/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 801.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[_Abel]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 2 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} 2 x b_{4}+y b_{5}+b_{2}+\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2}-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right )^{2} {\mathrm e}^{\frac {x^{2}}{2}} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{4}-\left (\frac {\left (-\frac {y \,x^{2} {\mathrm e}^{-\frac {x^{2}}{4}}}{2}+y \,{\mathrm e}^{-\frac {x^{2}}{4}}-2 y^{2} x \,{\mathrm e}^{-\frac {x^{2}}{2}}-3 y^{3} x \,{\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}+\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) x \,{\mathrm e}^{\frac {x^{2}}{4}}}{4}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (x \,{\mathrm e}^{-\frac {x^{2}}{4}}+4 y \,{\mathrm e}^{-\frac {x^{2}}{2}}+6 y^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}} \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{2} = 0 \end{equation} Putting the above in normal form gives \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Simplifying the above gives \begin{equation} \tag{6E} -2 x \,y^{3} a_{3} {\mathrm e}^{-\frac {x^{2}}{4}}-8 x \,y^{5} a_{5} {\mathrm e}^{-\frac {3 x^{2}}{4}}-4 x \,y^{5} a_{6} {\mathrm e}^{-\frac {x^{2}}{2}}-2 x^{2} y^{3} a_{5} {\mathrm e}^{-\frac {x^{2}}{4}}-6 x \,y^{4} a_{6} {\mathrm e}^{-\frac {x^{2}}{4}}+2 x^{3} y^{2} a_{4} {\mathrm e}^{-\frac {x^{2}}{4}}+4 b_{2}-6 \,{\mathrm e}^{\frac {x^{2}}{4}} x^{2} y a_{5}-10 \,{\mathrm e}^{\frac {x^{2}}{4}} x \,y^{2} a_{6}+4 x^{3} y^{3} a_{4} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x \,y^{3} a_{4} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x \,y^{3} b_{5} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x \,y^{2} a_{4} {\mathrm e}^{-\frac {x^{2}}{4}}-4 x \,y^{2} b_{5} {\mathrm e}^{-\frac {x^{2}}{4}}-4 x \,y^{6} a_{5} {\mathrm e}^{-x^{2}}-6 \,{\mathrm e}^{\frac {x^{2}}{4}} x y a_{3}-4 x \,y^{4} a_{5} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x \,y^{3} a_{5} {\mathrm e}^{-\frac {x^{2}}{4}}-12 x^{2} y^{2} b_{4} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x^{2} y b_{4} {\mathrm e}^{-\frac {x^{2}}{4}}+2 x^{2} y^{2} a_{2} {\mathrm e}^{-\frac {x^{2}}{4}}+2 x \,y^{2} a_{1} {\mathrm e}^{-\frac {x^{2}}{4}}+4 x^{2} y^{3} a_{2} {\mathrm e}^{-\frac {x^{2}}{2}}+4 x \,y^{3} a_{1} {\mathrm e}^{-\frac {x^{2}}{2}}-12 x \,y^{2} b_{2} {\mathrm e}^{-\frac {x^{2}}{2}}-8 x y b_{2} {\mathrm e}^{-\frac {x^{2}}{4}}-2 \,{\mathrm e}^{\frac {x^{2}}{4}} x^{3} a_{4}-8 \,{\mathrm e}^{\frac {x^{2}}{4}} x a_{4}+4 \,{\mathrm e}^{\frac {x^{2}}{4}} x b_{5}-4 \,{\mathrm e}^{\frac {x^{2}}{4}} y a_{5}+8 \,{\mathrm e}^{\frac {x^{2}}{4}} y b_{6}-6 x^{2} y a_{4}+8 x b_{4}+4 y b_{5}-4 \,{\mathrm e}^{\frac {x^{2}}{2}} a_{3}-4 \,{\mathrm e}^{\frac {x^{2}}{4}} a_{2}+4 \,{\mathrm e}^{\frac {x^{2}}{4}} b_{3}-18 y^{3} a_{6}-2 x^{3} b_{4}-10 y^{2} a_{3}-2 x^{2} b_{2}-2 x b_{1}-2 y a_{1}-12 x \,y^{2} a_{5}-4 y^{4} a_{5} {\mathrm e}^{-\frac {x^{2}}{2}}-4 y^{4} b_{6} {\mathrm e}^{-\frac {x^{2}}{2}}-4 y^{3} a_{5} {\mathrm e}^{-\frac {x^{2}}{4}}-8 y^{7} a_{6} {\mathrm e}^{-x^{2}}-8 y^{5} a_{6} {\mathrm e}^{-\frac {x^{2}}{2}}-16 y^{4} a_{6} {\mathrm e}^{-\frac {x^{2}}{4}}-4 \,{\mathrm e}^{\frac {x^{2}}{2}} x a_{5}-8 \,{\mathrm e}^{\frac {x^{2}}{2}} y a_{6}+2 x \,y^{2} b_{6}-16 y^{6} a_{6} {\mathrm e}^{-\frac {3 x^{2}}{4}}-x^{3} y^{2} a_{5}-2 x^{2} y^{3} a_{6}-2 \,{\mathrm e}^{\frac {x^{2}}{4}} x^{2} a_{2}-2 \,{\mathrm e}^{\frac {x^{2}}{4}} x a_{1}-8 y^{5} a_{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}-x^{2} y^{2} a_{3}-4 y^{2} a_{2} {\mathrm e}^{-\frac {x^{2}}{4}}-4 y^{2} b_{3} {\mathrm e}^{-\frac {x^{2}}{4}}-4 y^{6} a_{3} {\mathrm e}^{-x^{2}}-4 x y a_{2}-12 y^{2} b_{1} {\mathrm e}^{-\frac {x^{2}}{2}}-8 y b_{1} {\mathrm e}^{-\frac {x^{2}}{4}}-4 y^{3} a_{2} {\mathrm e}^{-\frac {x^{2}}{2}}-8 y^{3} b_{3} {\mathrm e}^{-\frac {x^{2}}{2}}-4 y^{4} a_{3} {\mathrm e}^{-\frac {x^{2}}{2}}-8 y^{3} a_{3} {\mathrm e}^{-\frac {x^{2}}{4}} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, {\mathrm e}^{-x^{2}}, {\mathrm e}^{-\frac {3 x^{2}}{4}}, {\mathrm e}^{-\frac {x^{2}}{2}}, {\mathrm e}^{-\frac {x^{2}}{4}}, {\mathrm e}^{\frac {x^{2}}{2}}, {\mathrm e}^{\frac {x^{2}}{4}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\mathrm e}^{-x^{2}} = v_{3}, {\mathrm e}^{-\frac {3 x^{2}}{4}} = v_{4}, {\mathrm e}^{-\frac {x^{2}}{2}} = v_{5}, {\mathrm e}^{-\frac {x^{2}}{4}} = v_{6}, {\mathrm e}^{\frac {x^{2}}{2}} = v_{7}, {\mathrm e}^{\frac {x^{2}}{4}} = v_{8}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -4 v_{1} v_{2}^{6} a_{5} v_{3}-8 v_{2}^{7} a_{6} v_{3}-4 v_{2}^{6} a_{3} v_{3}+4 v_{1}^{3} v_{2}^{3} a_{4} v_{5}-8 v_{1} v_{2}^{5} a_{5} v_{4}-4 v_{1} v_{2}^{5} a_{6} v_{5}-16 v_{2}^{6} a_{6} v_{4}+4 v_{1}^{2} v_{2}^{3} a_{2} v_{5}-8 v_{2}^{5} a_{3} v_{4}+2 v_{1}^{3} v_{2}^{2} a_{4} v_{6}-2 v_{1}^{2} v_{2}^{3} a_{5} v_{6}-4 v_{1} v_{2}^{4} a_{5} v_{5}-6 v_{1} v_{2}^{4} a_{6} v_{6}-8 v_{2}^{5} a_{6} v_{5}+4 v_{1} v_{2}^{3} a_{1} v_{5}+2 v_{1}^{2} v_{2}^{2} a_{2} v_{6}-2 v_{1} v_{2}^{3} a_{3} v_{6}-4 v_{2}^{4} a_{3} v_{5}-8 v_{1} v_{2}^{3} a_{4} v_{5}-v_{1}^{3} v_{2}^{2} a_{5}-8 v_{1} v_{2}^{3} a_{5} v_{6}-4 v_{2}^{4} a_{5} v_{5}-2 v_{1}^{2} v_{2}^{3} a_{6}-16 v_{2}^{4} a_{6} v_{6}-12 v_{1}^{2} v_{2}^{2} b_{4} v_{5}-8 v_{1} v_{2}^{3} b_{5} v_{5}-4 v_{2}^{4} b_{6} v_{5}+2 v_{1} v_{2}^{2} a_{1} v_{6}-4 v_{2}^{3} a_{2} v_{5}-v_{1}^{2} v_{2}^{2} a_{3}-8 v_{2}^{3} a_{3} v_{6}-2 v_{8} v_{1}^{3} a_{4}-8 v_{1} v_{2}^{2} a_{4} v_{6}-6 v_{8} v_{1}^{2} v_{2} a_{5}-4 v_{2}^{3} a_{5} v_{6}-10 v_{8} v_{1} v_{2}^{2} a_{6}-12 v_{1} v_{2}^{2} b_{2} v_{5}-8 v_{2}^{3} b_{3} v_{5}-8 v_{1}^{2} v_{2} b_{4} v_{6}-4 v_{1} v_{2}^{2} b_{5} v_{6}-2 v_{8} v_{1}^{2} a_{2}-4 v_{2}^{2} a_{2} v_{6}-6 v_{8} v_{1} v_{2} a_{3}-6 v_{1}^{2} v_{2} a_{4}-12 v_{1} v_{2}^{2} a_{5}-18 v_{2}^{3} a_{6}-12 v_{2}^{2} b_{1} v_{5}-8 v_{1} v_{2} b_{2} v_{6}-4 v_{2}^{2} b_{3} v_{6}-2 v_{1}^{3} b_{4}+2 v_{1} v_{2}^{2} b_{6}-2 v_{8} v_{1} a_{1}-4 v_{1} v_{2} a_{2}-10 v_{2}^{2} a_{3}-8 v_{8} v_{1} a_{4}-4 v_{7} v_{1} a_{5}-4 v_{8} v_{2} a_{5}-8 v_{7} v_{2} a_{6}-8 v_{2} b_{1} v_{6}-2 v_{1}^{2} b_{2}+4 v_{8} v_{1} b_{5}+8 v_{8} v_{2} b_{6}-2 v_{2} a_{1}-4 v_{8} a_{2}-4 v_{7} a_{3}-2 v_{1} b_{1}+4 v_{8} b_{3}+8 v_{1} b_{4}+4 v_{2} b_{5}+4 b_{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}, v_{8}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 4 b_{2}+\left (-12 a_{5}+2 b_{6}\right ) v_{1} v_{2}^{2}+\left (-2 a_{1}-8 a_{4}+4 b_{5}\right ) v_{1} v_{8}+\left (-4 a_{3}-4 a_{5}-4 b_{6}\right ) v_{2}^{4} v_{5}+\left (-4 a_{2}-8 b_{3}\right ) v_{2}^{3} v_{5}+\left (-8 a_{3}-4 a_{5}\right ) v_{2}^{3} v_{6}+\left (-4 a_{2}-4 b_{3}\right ) v_{2}^{2} v_{6}-8 v_{1} v_{2}^{5} a_{5} v_{4}-4 v_{1} v_{2}^{5} a_{6} v_{5}-2 v_{1}^{2} v_{2}^{3} a_{5} v_{6}-6 v_{1} v_{2}^{4} a_{6} v_{6}+2 v_{1}^{3} v_{2}^{2} a_{4} v_{6}-6 v_{8} v_{1}^{2} v_{2} a_{5}-10 v_{8} v_{1} v_{2}^{2} a_{6}+4 v_{1}^{3} v_{2}^{3} a_{4} v_{5}-4 v_{1} v_{2}^{6} a_{5} v_{3}-6 v_{8} v_{1} v_{2} a_{3}-4 v_{1} v_{2}^{4} a_{5} v_{5}-12 v_{1}^{2} v_{2}^{2} b_{4} v_{5}-8 v_{1}^{2} v_{2} b_{4} v_{6}+2 v_{1}^{2} v_{2}^{2} a_{2} v_{6}+4 v_{1}^{2} v_{2}^{3} a_{2} v_{5}-12 v_{1} v_{2}^{2} b_{2} v_{5}-8 v_{1} v_{2} b_{2} v_{6}+\left (-4 a_{5}+8 b_{6}\right ) v_{2} v_{8}-8 v_{2}^{7} a_{6} v_{3}-8 v_{2}^{5} a_{6} v_{5}-16 v_{2}^{4} a_{6} v_{6}-4 v_{7} v_{1} a_{5}-8 v_{7} v_{2} a_{6}-16 v_{2}^{6} a_{6} v_{4}-v_{1}^{3} v_{2}^{2} a_{5}-2 v_{1}^{2} v_{2}^{3} a_{6}-2 v_{8} v_{1}^{2} a_{2}-8 v_{2}^{5} a_{3} v_{4}-v_{1}^{2} v_{2}^{2} a_{3}-4 v_{2}^{6} a_{3} v_{3}-4 v_{1} v_{2} a_{2}+\left (4 a_{1}-8 a_{4}-8 b_{5}\right ) v_{1} v_{2}^{3} v_{5}+\left (-2 a_{3}-8 a_{5}\right ) v_{1} v_{2}^{3} v_{6}+\left (2 a_{1}-8 a_{4}-4 b_{5}\right ) v_{1} v_{2}^{2} v_{6}-2 v_{8} v_{1}^{3} a_{4}-6 v_{1}^{2} v_{2} a_{4}-4 v_{7} a_{3}-18 v_{2}^{3} a_{6}-2 v_{1}^{3} b_{4}-10 v_{2}^{2} a_{3}-2 v_{1}^{2} b_{2}-12 v_{2}^{2} b_{1} v_{5}-8 v_{2} b_{1} v_{6}+\left (-2 a_{1}+4 b_{5}\right ) v_{2}+\left (-4 a_{2}+4 b_{3}\right ) v_{8}+\left (-2 b_{1}+8 b_{4}\right ) v_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -4 a_{2}&=0\\ -2 a_{2}&=0\\ 2 a_{2}&=0\\ 4 a_{2}&=0\\ -10 a_{3}&=0\\ -8 a_{3}&=0\\ -6 a_{3}&=0\\ -4 a_{3}&=0\\ -a_{3}&=0\\ -6 a_{4}&=0\\ -2 a_{4}&=0\\ 2 a_{4}&=0\\ 4 a_{4}&=0\\ -8 a_{5}&=0\\ -6 a_{5}&=0\\ -4 a_{5}&=0\\ -2 a_{5}&=0\\ -a_{5}&=0\\ -18 a_{6}&=0\\ -16 a_{6}&=0\\ -10 a_{6}&=0\\ -8 a_{6}&=0\\ -6 a_{6}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ -12 b_{1}&=0\\ -8 b_{1}&=0\\ -12 b_{2}&=0\\ -8 b_{2}&=0\\ -2 b_{2}&=0\\ 4 b_{2}&=0\\ -12 b_{4}&=0\\ -8 b_{4}&=0\\ -2 b_{4}&=0\\ -2 a_{1}+4 b_{5}&=0\\ -4 a_{2}-8 b_{3}&=0\\ -4 a_{2}-4 b_{3}&=0\\ -4 a_{2}+4 b_{3}&=0\\ -8 a_{3}-4 a_{5}&=0\\ -2 a_{3}-8 a_{5}&=0\\ -12 a_{5}+2 b_{6}&=0\\ -4 a_{5}+8 b_{6}&=0\\ -2 b_{1}+8 b_{4}&=0\\ -2 a_{1}-8 a_{4}+4 b_{5}&=0\\ 2 a_{1}-8 a_{4}-4 b_{5}&=0\\ 4 a_{1}-8 a_{4}-8 b_{5}&=0\\ -4 a_{3}-4 a_{5}-4 b_{6}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=2 b_{5}\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=b_{5}\\ b_{6}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 2 \\ \eta &= y x \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}}+y^{2} {\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}+\frac {y \,{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} x}{2}+{\mathrm e}^{\frac {x^{2}}{4}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= {\mathrm e}^{\frac {x^{2}}{4}}\\ f_1(x) &= \frac {{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} x}{2}\\ f_2(x) &= {\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}\\ f_3(x) &= {\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {3 x^{2}}{4}} \end {align*}
Since \(f_2(x)={\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {{\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}}{3 \,{\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {3 x^{2}}{4}}} \right ) \\ &= u \left (x \right )-\frac {{\mathrm e}^{\frac {x^{2}}{4}}}{3} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = {\mathrm e}^{-\frac {3 x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} u \left (x \right )^{3}+\frac {{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} u \left (x \right ) x}{2}-\frac {{\mathrm e}^{\frac {3 x^{2}}{4}} {\mathrm e}^{-x^{2}} {\mathrm e}^{\frac {x^{2}}{4}} u \left (x \right )}{3}-\frac {{\mathrm e}^{\frac {3 x^{2}}{4}} {\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} x}{6}+\frac {2 \,{\mathrm e}^{\frac {3 x^{2}}{2}} {\mathrm e}^{-\frac {3 x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}}{27}+\frac {x \,{\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {3 x^{2}}{4}}}{6}+{\mathrm e}^{\frac {x^{2}}{4}}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Writing the ode as \begin {align*} u^{\prime }\left (x \right )&=\frac {{\mathrm e}^{\frac {3 x^{2}}{2}} \left (54 \,{\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {9 x^{2}}{4}} u^{3}+27 \,{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {3 x^{2}}{2}} u x -18 \,{\mathrm e}^{-x^{2}} {\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {3 x^{2}}{4}} u -9 \,{\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} {\mathrm e}^{-\frac {3 x^{2}}{4}} x +58 \,{\mathrm e}^{-\frac {3 x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}}+9 \,{\mathrm e}^{-\frac {x^{2}}{2}} x \,{\mathrm e}^{-\frac {3 x^{2}}{4}}\right )}{54}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{u}-\xi _{x}\right ) -\omega ^{2}\xi _{u}-\omega _{x}\xi -\omega _{u}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 2 to use as anstaz gives \begin{align*} \tag{1E} \xi &= u^{2} a_{6}+u x a_{5}+x^{2} a_{4}+u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u^{2} b_{6}+u x b_{5}+x^{2} b_{4}+u b_{3}+x b_{2}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} \text {Expression too large to display} \end{equation} Putting the above in normal form gives \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Looking at the above PDE shows the following are all the terms with \(\{u, x\}\) in them. \[ \left \{u, x, {\mathrm e}^{-4 x^{2}}, {\mathrm e}^{-3 x^{2}}, {\mathrm e}^{-\frac {13 x^{2}}{4}}, {\mathrm e}^{-\frac {11 x^{2}}{4}}, {\mathrm e}^{-\frac {7 x^{2}}{2}}, {\mathrm e}^{-\frac {5 x^{2}}{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{u, x\}\) in them \[ \left \{u = v_{1}, x = v_{2}, {\mathrm e}^{-4 x^{2}} = v_{3}, {\mathrm e}^{-3 x^{2}} = v_{4}, {\mathrm e}^{-\frac {13 x^{2}}{4}} = v_{5}, {\mathrm e}^{-\frac {11 x^{2}}{4}} = v_{6}, {\mathrm e}^{-\frac {7 x^{2}}{2}} = v_{7}, {\mathrm e}^{-\frac {5 x^{2}}{2}} = v_{8}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -2916 v_{1}^{6} v_{2} a_{5} v_{3}-5832 v_{1}^{7} a_{6} v_{3}-2916 v_{1}^{6} a_{3} v_{3}+2916 v_{1}^{3} v_{2}^{3} a_{4} v_{7}-2916 v_{1}^{5} v_{2} a_{6} v_{7}+2916 v_{1}^{3} v_{2}^{2} a_{2} v_{7}+1944 v_{1}^{4} v_{2} a_{5} v_{7}-729 v_{1}^{2} v_{2}^{3} a_{5} v_{4}+3888 v_{1}^{5} a_{6} v_{7}-1458 v_{1}^{3} v_{2}^{2} a_{6} v_{4}+2916 v_{1}^{3} v_{2} a_{1} v_{7}+1944 v_{1}^{4} a_{3} v_{7}-729 v_{1}^{2} v_{2}^{2} a_{3} v_{4}-5832 v_{1}^{3} v_{2} a_{4} v_{7}-2916 v_{1}^{4} a_{5} v_{7}-6264 v_{1}^{3} v_{2} a_{5} v_{5}+972 v_{1}^{2} v_{2}^{2} a_{5} v_{4}-12528 v_{1}^{4} a_{6} v_{5}+1944 v_{1}^{3} v_{2} a_{6} v_{4}-8748 v_{1}^{2} v_{2}^{2} b_{4} v_{7}-5832 v_{1}^{3} v_{2} b_{5} v_{7}-2916 v_{1}^{4} b_{6} v_{7}-2916 v_{1}^{3} a_{2} v_{7}-6264 v_{1}^{3} a_{3} v_{5}+972 v_{1}^{2} v_{2} a_{3} v_{4}-4374 v_{1} v_{2}^{2} a_{4} v_{4}-1566 v_{2}^{3} a_{4} v_{6}-3240 v_{1}^{2} v_{2} a_{5} v_{4}-4698 v_{1} v_{2}^{2} a_{5} v_{6}-2106 v_{1}^{3} a_{6} v_{4}-7830 v_{1}^{2} v_{2} a_{6} v_{6}-8748 v_{1}^{2} v_{2} b_{2} v_{7}-5832 v_{1}^{3} b_{3} v_{7}-1458 v_{2}^{3} b_{4} v_{4}+1458 v_{1}^{2} v_{2} b_{6} v_{4}-2916 v_{1} v_{2} a_{2} v_{4}-1566 v_{2}^{2} a_{2} v_{6}-1782 v_{1}^{2} a_{3} v_{4}-4698 v_{1} v_{2} a_{3} v_{6}+1944 v_{1} v_{2} a_{4} v_{4}+972 v_{1}^{2} a_{5} v_{4}+2088 v_{1} v_{2} a_{5} v_{6}+4176 v_{1}^{2} a_{6} v_{6}-8748 v_{1}^{2} b_{1} v_{7}-1458 v_{2}^{2} b_{2} v_{4}+972 v_{2}^{2} b_{4} v_{4}-972 v_{1}^{2} b_{6} v_{4}-1458 v_{1} a_{1} v_{4}-1566 v_{2} a_{1} v_{6}+972 v_{1} a_{2} v_{4}+2088 v_{1} a_{3} v_{6}-6264 v_{2} a_{4} v_{6}-3132 v_{1} a_{5} v_{6}-3364 v_{2} a_{5} v_{8}-6728 v_{1} a_{6} v_{8}-1458 v_{2} b_{1} v_{4}+972 v_{2} b_{2} v_{4}+5832 v_{2} b_{4} v_{4}+2916 v_{1} b_{5} v_{4}+3132 v_{2} b_{5} v_{6}+6264 v_{1} b_{6} v_{6}-3132 a_{2} v_{6}-3364 a_{3} v_{8}+972 b_{1} v_{4}+2916 b_{2} v_{4}+3132 b_{3} v_{6} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, v_{7}, v_{8}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (2916 a_{1}-5832 a_{4}-5832 b_{5}\right ) v_{1}^{3} v_{2} v_{7}+\left (-729 a_{3}+972 a_{5}\right ) v_{1}^{2} v_{2}^{2} v_{4}+\left (972 a_{3}-3240 a_{5}+1458 b_{6}\right ) v_{1}^{2} v_{2} v_{4}+\left (-2916 a_{2}+1944 a_{4}\right ) v_{1} v_{2} v_{4}+\left (-4698 a_{3}+2088 a_{5}\right ) v_{1} v_{2} v_{6}-2916 v_{1}^{6} a_{3} v_{3}-6264 v_{1}^{3} a_{3} v_{5}-8748 v_{1}^{2} b_{1} v_{7}-1566 v_{2}^{2} a_{2} v_{6}-5832 v_{1}^{7} a_{6} v_{3}-12528 v_{1}^{4} a_{6} v_{5}-6728 v_{1} a_{6} v_{8}-3364 v_{2} a_{5} v_{8}-1566 v_{2}^{3} a_{4} v_{6}+4176 v_{1}^{2} a_{6} v_{6}-2106 v_{1}^{3} a_{6} v_{4}-1458 v_{2}^{3} b_{4} v_{4}+3888 v_{1}^{5} a_{6} v_{7}+\left (-2916 a_{2}-5832 b_{3}\right ) v_{1}^{3} v_{7}+\left (-1782 a_{3}+972 a_{5}-972 b_{6}\right ) v_{1}^{2} v_{4}+\left (-1458 a_{1}+972 a_{2}+2916 b_{5}\right ) v_{1} v_{4}+\left (2088 a_{3}-3132 a_{5}+6264 b_{6}\right ) v_{1} v_{6}+\left (-1458 b_{2}+972 b_{4}\right ) v_{2}^{2} v_{4}+\left (-1458 b_{1}+972 b_{2}+5832 b_{4}\right ) v_{2} v_{4}+\left (-1566 a_{1}-6264 a_{4}+3132 b_{5}\right ) v_{2} v_{6}-3364 a_{3} v_{8}-2916 v_{1}^{6} v_{2} a_{5} v_{3}+2916 v_{1}^{3} v_{2}^{3} a_{4} v_{7}+2916 v_{1}^{3} v_{2}^{2} a_{2} v_{7}-8748 v_{1}^{2} v_{2} b_{2} v_{7}-8748 v_{1}^{2} v_{2}^{2} b_{4} v_{7}-7830 v_{1}^{2} v_{2} a_{6} v_{6}-4698 v_{1} v_{2}^{2} a_{5} v_{6}+1944 v_{1}^{3} v_{2} a_{6} v_{4}-2916 v_{1}^{5} v_{2} a_{6} v_{7}+1944 v_{1}^{4} v_{2} a_{5} v_{7}-1458 v_{1}^{3} v_{2}^{2} a_{6} v_{4}-729 v_{1}^{2} v_{2}^{3} a_{5} v_{4}-6264 v_{1}^{3} v_{2} a_{5} v_{5}-4374 v_{1} v_{2}^{2} a_{4} v_{4}+\left (1944 a_{3}-2916 a_{5}-2916 b_{6}\right ) v_{1}^{4} v_{7}+\left (972 b_{1}+2916 b_{2}\right ) v_{4}+\left (-3132 a_{2}+3132 b_{3}\right ) v_{6} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -1566 a_{2}&=0\\ 2916 a_{2}&=0\\ -6264 a_{3}&=0\\ -3364 a_{3}&=0\\ -2916 a_{3}&=0\\ -4374 a_{4}&=0\\ -1566 a_{4}&=0\\ 2916 a_{4}&=0\\ -6264 a_{5}&=0\\ -4698 a_{5}&=0\\ -3364 a_{5}&=0\\ -2916 a_{5}&=0\\ -729 a_{5}&=0\\ 1944 a_{5}&=0\\ -12528 a_{6}&=0\\ -7830 a_{6}&=0\\ -6728 a_{6}&=0\\ -5832 a_{6}&=0\\ -2916 a_{6}&=0\\ -2106 a_{6}&=0\\ -1458 a_{6}&=0\\ 1944 a_{6}&=0\\ 3888 a_{6}&=0\\ 4176 a_{6}&=0\\ -8748 b_{1}&=0\\ -8748 b_{2}&=0\\ -8748 b_{4}&=0\\ -1458 b_{4}&=0\\ -3132 a_{2}+3132 b_{3}&=0\\ -2916 a_{2}+1944 a_{4}&=0\\ -2916 a_{2}-5832 b_{3}&=0\\ -4698 a_{3}+2088 a_{5}&=0\\ -729 a_{3}+972 a_{5}&=0\\ 972 b_{1}+2916 b_{2}&=0\\ -1458 b_{2}+972 b_{4}&=0\\ -1566 a_{1}-6264 a_{4}+3132 b_{5}&=0\\ -1458 a_{1}+972 a_{2}+2916 b_{5}&=0\\ 2916 a_{1}-5832 a_{4}-5832 b_{5}&=0\\ -1782 a_{3}+972 a_{5}-972 b_{6}&=0\\ 972 a_{3}-3240 a_{5}+1458 b_{6}&=0\\ 1944 a_{3}-2916 a_{5}-2916 b_{6}&=0\\ 2088 a_{3}-3132 a_{5}+6264 b_{6}&=0\\ -1458 b_{1}+972 b_{2}+5832 b_{4}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=2 b_{5}\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=b_{5}\\ b_{6}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 2 \\ \eta &= u x \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Since unable to solve for \(u \left (x \right )\), will terminate solution.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (y \,{\mathrm e}^{-\frac {x^{2}}{4}} x +2+2 y^{2} {\mathrm e}^{-\frac {x^{2}}{2}}+2 y^{3} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 45
dsolve(diff(y(x),x) = 1/2*(y(x)*exp(-1/4*x^2)*x+2+2*y(x)^2*exp(-1/2*x^2)+2*y(x)^3*exp(-3/4*x^2))*exp(1/4*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {29 \,{\mathrm e}^{\frac {x^{2}}{4}} \operatorname {RootOf}\left (-81 \left (\int _{}^{\textit {\_Z}}\frac {1}{841 \textit {\_a}^{3}-27 \textit {\_a} +27}d \textit {\_a} \right )+x +3 c_{1} \right )}{9}-\frac {{\mathrm e}^{\frac {x^{2}}{4}}}{3} \]
✓ Solution by Mathematica
Time used: 0.322 (sec). Leaf size: 126
DSolve[y'[x] == (E^(x^2/4)*(2 + (x*y[x])/E^(x^2/4) + (2*y[x]^2)/E^(x^2/2) + (2*y[x]^3)/E^((3*x^2)/4)))/2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {29}{3} \text {RootSum}\left [-29 \text {$\#$1}^3+3 \sqrt [3]{29} \text {$\#$1}-29\&,\frac {\log \left (\frac {3 e^{-\frac {x^2}{2}} y(x)+e^{-\frac {x^2}{4}}}{\sqrt [3]{29} \sqrt [3]{e^{-\frac {3 x^2}{4}}}}-\text {$\#$1}\right )}{\sqrt [3]{29}-29 \text {$\#$1}^2}\&\right ]=\frac {1}{9} 29^{2/3} e^{\frac {x^2}{2}} \left (e^{-\frac {3 x^2}{4}}\right )^{2/3} x+c_1,y(x)\right ] \]