Internal problem ID [9294]
Internal file name [OUTPUT/8230_Monday_June_06_2022_02_24_13_AM_31650770/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 961.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{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 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right ) \left (b_{3}-a_{2}\right )}{y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}-\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )^{2} a_{3}}{\left (y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )^{2}}-\left (\frac {2 y +2 x +\left (-12 x^{5}+24 y^{2} x^{3}-12 y^{4} x +8 x^{3}-8 y^{2} x \right ) {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}-\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right ) \left (2 y +2 x -\left (-12 x^{5}+24 y^{2} x^{3}-12 y^{4} x +8 x^{3}-8 y^{2} x \right ) {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )}{\left (y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2 y +2 x +\left (12 x^{4} y -24 y^{3} x^{2}+12 y^{5}-8 x^{2} y +8 y^{3}\right ) {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}}-\frac {\left (y^{2}+2 y x +x^{2}+{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right ) \left (2 y +2 x -\left (12 x^{4} y -24 y^{3} x^{2}+12 y^{5}-8 x^{2} y +8 y^{3}\right ) {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )}{\left (y^{2}+2 y x +x^{2}-{\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 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} \text {Expression too large to display} \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y, {\mathrm e}^{-4 x^{6}+12 y^{2} x^{4}-12 y^{4} x^{2}+4 y^{6}+4 x^{4}-8 y^{2} x^{2}+4 y^{4}+4}, {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2}\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}, {\mathrm e}^{-4 x^{6}+12 y^{2} x^{4}-12 y^{4} x^{2}+4 y^{6}+4 x^{4}-8 y^{2} x^{2}+4 y^{4}+4} = v_{3}, {\mathrm e}^{-2 x^{6}+6 y^{2} x^{4}-6 y^{4} x^{2}+2 y^{6}+2 x^{4}-4 y^{2} x^{2}+2 y^{4}+2} = v_{4}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 24 v_{4} v_{1}^{8} a_{2}+48 v_{4} v_{1}^{7} v_{2} a_{2}-24 v_{4} v_{1}^{6} v_{2}^{2} a_{2}-96 v_{4} v_{1}^{5} v_{2}^{3} a_{2}-24 v_{4} v_{1}^{4} v_{2}^{4} a_{2}+48 v_{4} v_{1}^{3} v_{2}^{5} a_{2}+24 v_{4} v_{1}^{2} v_{2}^{6} a_{2}+24 v_{4} v_{1}^{7} v_{2} a_{3}+48 v_{4} v_{1}^{6} v_{2}^{2} a_{3}-24 v_{4} v_{1}^{5} v_{2}^{3} a_{3}-96 v_{4} v_{1}^{4} v_{2}^{4} a_{3}-24 v_{4} v_{1}^{3} v_{2}^{5} a_{3}+48 v_{4} v_{1}^{2} v_{2}^{6} a_{3}+24 v_{4} v_{1} v_{2}^{7} a_{3}-24 v_{4} v_{1}^{7} v_{2} b_{2}-48 v_{4} v_{1}^{6} v_{2}^{2} b_{2}+24 v_{4} v_{1}^{5} v_{2}^{3} b_{2}+96 v_{4} v_{1}^{4} v_{2}^{4} b_{2}+24 v_{4} v_{1}^{3} v_{2}^{5} b_{2}-48 v_{4} v_{1}^{2} v_{2}^{6} b_{2}-24 v_{4} v_{1} v_{2}^{7} b_{2}-24 v_{4} v_{1}^{6} v_{2}^{2} b_{3}-48 v_{4} v_{1}^{5} v_{2}^{3} b_{3}+24 v_{4} v_{1}^{4} v_{2}^{4} b_{3}+96 v_{4} v_{1}^{3} v_{2}^{5} b_{3}+24 v_{4} v_{1}^{2} v_{2}^{6} b_{3}-48 v_{4} v_{1} v_{2}^{7} b_{3}-24 v_{4} v_{2}^{8} b_{3}+24 v_{4} v_{1}^{7} a_{1}+48 v_{4} v_{1}^{6} v_{2} a_{1}-24 v_{4} v_{1}^{5} v_{2}^{2} a_{1}-96 v_{4} v_{1}^{4} v_{2}^{3} a_{1}-24 v_{4} v_{1}^{3} v_{2}^{4} a_{1}+48 v_{4} v_{1}^{2} v_{2}^{5} a_{1}+24 v_{4} v_{1} v_{2}^{6} a_{1}-24 v_{4} v_{1}^{6} v_{2} b_{1}-48 v_{4} v_{1}^{5} v_{2}^{2} b_{1}+24 v_{4} v_{1}^{4} v_{2}^{3} b_{1}+96 v_{4} v_{1}^{3} v_{2}^{4} b_{1}+24 v_{4} v_{1}^{2} v_{2}^{5} b_{1}-48 v_{4} v_{1} v_{2}^{6} b_{1}-24 v_{4} v_{2}^{7} b_{1}-16 v_{4} v_{1}^{6} a_{2}-32 v_{4} v_{1}^{5} v_{2} a_{2}+32 v_{4} v_{1}^{3} v_{2}^{3} a_{2}+16 v_{4} v_{1}^{2} v_{2}^{4} a_{2}-16 v_{4} v_{1}^{5} v_{2} a_{3}-32 v_{4} v_{1}^{4} v_{2}^{2} a_{3}+32 v_{4} v_{1}^{2} v_{2}^{4} a_{3}+16 v_{4} v_{1} v_{2}^{5} a_{3}+16 v_{4} v_{1}^{5} v_{2} b_{2}+32 v_{4} v_{1}^{4} v_{2}^{2} b_{2}-32 v_{4} v_{1}^{2} v_{2}^{4} b_{2}-16 v_{4} v_{1} v_{2}^{5} b_{2}+16 v_{4} v_{1}^{4} v_{2}^{2} b_{3}+32 v_{4} v_{1}^{3} v_{2}^{3} b_{3}-32 v_{4} v_{1} v_{2}^{5} b_{3}-16 v_{4} v_{2}^{6} b_{3}-16 v_{4} v_{1}^{5} a_{1}-32 v_{4} v_{1}^{4} v_{2} a_{1}+32 v_{4} v_{1}^{2} v_{2}^{3} a_{1}+16 v_{4} v_{1} v_{2}^{4} a_{1}+16 v_{4} v_{1}^{4} v_{2} b_{1}+32 v_{4} v_{1}^{3} v_{2}^{2} b_{1}-32 v_{4} v_{1} v_{2}^{4} b_{1}-16 v_{4} v_{2}^{5} b_{1}-v_{1}^{4} a_{2}-4 v_{1}^{3} v_{2} a_{2}-6 v_{1}^{2} v_{2}^{2} a_{2}-4 v_{1} v_{2}^{3} a_{2}-v_{2}^{4} a_{2}-v_{1}^{4} a_{3}-4 v_{1}^{3} v_{2} a_{3}-6 v_{1}^{2} v_{2}^{2} a_{3}-4 v_{1} v_{2}^{3} a_{3}-v_{2}^{4} a_{3}+v_{1}^{4} b_{2}+4 v_{1}^{3} v_{2} b_{2}+6 v_{1}^{2} v_{2}^{2} b_{2}+4 v_{1} v_{2}^{3} b_{2}+v_{2}^{4} b_{2}+v_{1}^{4} b_{3}+4 v_{1}^{3} v_{2} b_{3}+6 v_{1}^{2} v_{2}^{2} b_{3}+4 v_{1} v_{2}^{3} b_{3}+v_{2}^{4} b_{3}+4 v_{4} v_{1}^{2} a_{2}+4 v_{4} v_{1} v_{2} a_{2}-2 v_{4} v_{1}^{2} a_{3}+2 v_{4} v_{2}^{2} a_{3}+2 v_{4} v_{1}^{2} b_{2}-2 v_{4} v_{2}^{2} b_{2}+4 v_{4} v_{1} v_{2} b_{3}+4 v_{4} v_{2}^{2} b_{3}+4 v_{4} v_{1} a_{1}+4 v_{4} v_{2} a_{1}+4 v_{4} v_{1} b_{1}+4 v_{4} v_{2} b_{1}+v_{3} a_{2}-v_{3} a_{3}+v_{3} b_{2}-v_{3} b_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (4 a_{1}+4 b_{1}\right ) v_{2} v_{4}+\left (-4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}\right ) v_{1}^{3} v_{2}+\left (-6 a_{2}-6 a_{3}+6 b_{2}+6 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (4 a_{2}-2 a_{3}+2 b_{2}\right ) v_{1}^{2} v_{4}+\left (-4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}\right ) v_{1} v_{2}^{3}+\left (4 a_{1}+4 b_{1}\right ) v_{1} v_{4}+\left (2 a_{3}-2 b_{2}+4 b_{3}\right ) v_{2}^{2} v_{4}+32 v_{4} v_{1}^{3} v_{2}^{2} b_{1}+32 v_{4} v_{1}^{2} v_{2}^{3} a_{1}+\left (48 a_{2}+24 a_{3}-24 b_{2}\right ) v_{1}^{7} v_{2} v_{4}+\left (-24 a_{2}+48 a_{3}-48 b_{2}-24 b_{3}\right ) v_{1}^{6} v_{2}^{2} v_{4}+\left (48 a_{1}-24 b_{1}\right ) v_{1}^{6} v_{2} v_{4}+\left (-96 a_{2}-24 a_{3}+24 b_{2}-48 b_{3}\right ) v_{1}^{5} v_{2}^{3} v_{4}+\left (-24 a_{1}-48 b_{1}\right ) v_{1}^{5} v_{2}^{2} v_{4}+\left (-32 a_{2}-16 a_{3}+16 b_{2}\right ) v_{1}^{5} v_{2} v_{4}+\left (-24 a_{2}-96 a_{3}+96 b_{2}+24 b_{3}\right ) v_{1}^{4} v_{2}^{4} v_{4}+\left (-96 a_{1}+24 b_{1}\right ) v_{1}^{4} v_{2}^{3} v_{4}+\left (-32 a_{3}+32 b_{2}+16 b_{3}\right ) v_{1}^{4} v_{2}^{2} v_{4}+\left (-32 a_{1}+16 b_{1}\right ) v_{1}^{4} v_{2} v_{4}+\left (48 a_{2}-24 a_{3}+24 b_{2}+96 b_{3}\right ) v_{1}^{3} v_{2}^{5} v_{4}+\left (-24 a_{1}+96 b_{1}\right ) v_{1}^{3} v_{2}^{4} v_{4}+\left (32 a_{2}+32 b_{3}\right ) v_{1}^{3} v_{2}^{3} v_{4}+\left (24 a_{2}+48 a_{3}-48 b_{2}+24 b_{3}\right ) v_{1}^{2} v_{2}^{6} v_{4}+\left (48 a_{1}+24 b_{1}\right ) v_{1}^{2} v_{2}^{5} v_{4}+\left (16 a_{2}+32 a_{3}-32 b_{2}\right ) v_{1}^{2} v_{2}^{4} v_{4}+\left (24 a_{3}-24 b_{2}-48 b_{3}\right ) v_{1} v_{2}^{7} v_{4}+\left (24 a_{1}-48 b_{1}\right ) v_{1} v_{2}^{6} v_{4}+\left (16 a_{3}-16 b_{2}-32 b_{3}\right ) v_{1} v_{2}^{5} v_{4}+\left (16 a_{1}-32 b_{1}\right ) v_{1} v_{2}^{4} v_{4}+\left (4 a_{2}+4 b_{3}\right ) v_{1} v_{2} v_{4}+\left (-a_{2}-a_{3}+b_{2}+b_{3}\right ) v_{1}^{4}+\left (-a_{2}-a_{3}+b_{2}+b_{3}\right ) v_{2}^{4}+\left (a_{2}-a_{3}+b_{2}-b_{3}\right ) v_{3}+24 v_{4} v_{1}^{7} a_{1}-24 v_{4} v_{2}^{7} b_{1}-16 v_{4} v_{1}^{6} a_{2}-16 v_{4} v_{2}^{6} b_{3}-16 v_{4} v_{1}^{5} a_{1}-16 v_{4} v_{2}^{5} b_{1}+24 v_{4} v_{1}^{8} a_{2}-24 v_{4} v_{2}^{8} b_{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -16 a_{1}&=0\\ 24 a_{1}&=0\\ 32 a_{1}&=0\\ -16 a_{2}&=0\\ 24 a_{2}&=0\\ -24 b_{1}&=0\\ -16 b_{1}&=0\\ 32 b_{1}&=0\\ -24 b_{3}&=0\\ -16 b_{3}&=0\\ -96 a_{1}+24 b_{1}&=0\\ -32 a_{1}+16 b_{1}&=0\\ -24 a_{1}-48 b_{1}&=0\\ -24 a_{1}+96 b_{1}&=0\\ 4 a_{1}+4 b_{1}&=0\\ 16 a_{1}-32 b_{1}&=0\\ 24 a_{1}-48 b_{1}&=0\\ 48 a_{1}-24 b_{1}&=0\\ 48 a_{1}+24 b_{1}&=0\\ 4 a_{2}+4 b_{3}&=0\\ 32 a_{2}+32 b_{3}&=0\\ -32 a_{2}-16 a_{3}+16 b_{2}&=0\\ 4 a_{2}-2 a_{3}+2 b_{2}&=0\\ 16 a_{2}+32 a_{3}-32 b_{2}&=0\\ 48 a_{2}+24 a_{3}-24 b_{2}&=0\\ -32 a_{3}+32 b_{2}+16 b_{3}&=0\\ 2 a_{3}-2 b_{2}+4 b_{3}&=0\\ 16 a_{3}-16 b_{2}-32 b_{3}&=0\\ 24 a_{3}-24 b_{2}-48 b_{3}&=0\\ -96 a_{2}-24 a_{3}+24 b_{2}-48 b_{3}&=0\\ -24 a_{2}-96 a_{3}+96 b_{2}+24 b_{3}&=0\\ -24 a_{2}+48 a_{3}-48 b_{2}-24 b_{3}&=0\\ -6 a_{2}-6 a_{3}+6 b_{2}+6 b_{3}&=0\\ -4 a_{2}-4 a_{3}+4 b_{2}+4 b_{3}&=0\\ -a_{2}-a_{3}+b_{2}+b_{3}&=0\\ a_{2}-a_{3}+b_{2}-b_{3}&=0\\ 24 a_{2}+48 a_{3}-48 b_{2}+24 b_{3}&=0\\ 48 a_{2}-24 a_{3}+24 b_{2}+96 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=b_{2}\\ b_{1}&=0\\ b_{2}&=b_{2}\\ b_{3}&=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 &= y \\ \eta &= 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} y^{\prime }+2 y^{\prime } x y+x^{2} y^{\prime }-y^{\prime } {\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}-y^{2}-2 y x -x^{2}-{\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}}{y^{2}+2 y x +x^{2}-{\mathrm e}^{2+2 y^{4}-4 y^{2} x^{2}+2 x^{4}+2 y^{6}-6 y^{4} x^{2}+6 y^{2} x^{4}-2 x^{6}}} \end {array} \]
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 differential order: 1; found: 1 linear symmetries. Trying reduction of order 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 45
dsolve(diff(y(x),x) = (y(x)^2+2*x*y(x)+x^2+exp(2+2*y(x)^4-4*x^2*y(x)^2+2*x^4+2*y(x)^6-6*x^2*y(x)^4+6*x^4*y(x)^2-2*x^6))/(y(x)^2+2*x*y(x)+x^2-exp(2+2*y(x)^4-4*x^2*y(x)^2+2*x^4+2*y(x)^6-6*x^2*y(x)^4+6*x^4*y(x)^2-2*x^6)),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} +\int _{}^{{\mathrm e}^{2 \textit {\_Z}}-2 x \,{\mathrm e}^{\textit {\_Z}}}\frac {1}{{\mathrm e}^{2 \textit {\_a}^{3}+2 \textit {\_a}^{2}+2}+\textit {\_a}}d \textit {\_a} +c_{1} \right )}-x \]
✓ Solution by Mathematica
Time used: 29.082 (sec). Leaf size: 813
DSolve[y'[x] == (E^(2 + 2*x^4 - 2*x^6 - 4*x^2*y[x]^2 + 6*x^4*y[x]^2 + 2*y[x]^4 - 6*x^2*y[x]^4 + 2*y[x]^6) + x^2 + 2*x*y[x] + y[x]^2)/(-E^(2 + 2*x^4 - 2*x^6 - 4*x^2*y[x]^2 + 6*x^4*y[x]^2 + 2*y[x]^4 - 6*x^2*y[x]^4 + 2*y[x]^6) + x^2 + 2*x*y[x] + y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\left (\frac {1}{K[1]+y(x)}-\frac {2 e^{2 K[1]^6+6 y(x)^4 K[1]^2+4 y(x)^2 K[1]^2} K[1]}{e^{2 K[1]^6+6 y(x)^4 K[1]^2+4 y(x)^2 K[1]^2} K[1]^2-e^{2 y(x)^6+2 y(x)^4+6 K[1]^4 y(x)^2+2 K[1]^4+2}-e^{2 K[1]^6+6 y(x)^4 K[1]^2+4 y(x)^2 K[1]^2} y(x)^2}\right )dK[1]+\int _1^{y(x)}\left (-\frac {2 e^{2 x^6+6 K[2]^4 x^2+4 K[2]^2 x^2} K[2]}{-e^{2 x^6+6 K[2]^4 x^2+4 K[2]^2 x^2} x^2+e^{2 K[2]^6+2 K[2]^4+6 x^4 K[2]^2+2 x^4+2}+e^{2 x^6+6 K[2]^4 x^2+4 K[2]^2 x^2} K[2]^2}-\int _1^x\left (-\frac {2 e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[1] \left (24 K[1]^2 K[2]^3+8 K[1]^2 K[2]\right )}{e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[1]^2-e^{2 K[2]^6+2 K[2]^4+6 K[1]^4 K[2]^2+2 K[1]^4+2}-e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[2]^2}+\frac {2 e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[1] \left (e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} \left (24 K[1]^2 K[2]^3+8 K[1]^2 K[2]\right ) K[1]^2-2 e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[2]-e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[2]^2 \left (24 K[1]^2 K[2]^3+8 K[1]^2 K[2]\right )-e^{2 K[2]^6+2 K[2]^4+6 K[1]^4 K[2]^2+2 K[1]^4+2} \left (12 K[2]^5+8 K[2]^3+12 K[1]^4 K[2]\right )\right )}{\left (e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[1]^2-e^{2 K[2]^6+2 K[2]^4+6 K[1]^4 K[2]^2+2 K[1]^4+2}-e^{2 K[1]^6+6 K[2]^4 K[1]^2+4 K[2]^2 K[1]^2} K[2]^2\right )^2}-\frac {1}{(K[1]+K[2])^2}\right )dK[1]+\frac {1}{x+K[2]}\right )dK[2]=c_1,y(x)\right ] \]