Internal problem ID [9213]
Internal file name [OUTPUT/8149_Monday_June_06_2022_01_56_46_AM_20603524/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 880.
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.
\[ \boxed {y^{\prime }+\frac {2 a}{-y-2 a -2 a y^{4}+16 y^{2} a^{2} x -32 a^{3} x^{2}-2 y^{6} a +24 y^{4} a^{2} x -96 y^{2} a^{3} x^{2}+128 a^{4} x^{3}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {2 a}{-128 a^{4} x^{3}+96 y^{2} a^{3} x^{2}-24 a^{2} x \,y^{4}+2 a \,y^{6}+32 a^{3} x^{2}-16 y^{2} a^{2} x +2 a \,y^{4}+2 a +y}\\ 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 {2 a \left (b_{3}-a_{2}\right )}{-128 a^{4} x^{3}+96 y^{2} a^{3} x^{2}-24 a^{2} x \,y^{4}+2 a \,y^{6}+32 a^{3} x^{2}-16 y^{2} a^{2} x +2 a \,y^{4}+2 a +y}-\frac {4 a^{2} a_{3}}{\left (-128 a^{4} x^{3}+96 y^{2} a^{3} x^{2}-24 a^{2} x \,y^{4}+2 a \,y^{6}+32 a^{3} x^{2}-16 y^{2} a^{2} x +2 a \,y^{4}+2 a +y \right )^{2}}+\frac {2 a \left (-384 a^{4} x^{2}+192 a^{3} x \,y^{2}-24 a^{2} y^{4}+64 a^{3} x -16 a^{2} y^{2}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )}{\left (-128 a^{4} x^{3}+96 y^{2} a^{3} x^{2}-24 a^{2} x \,y^{4}+2 a \,y^{6}+32 a^{3} x^{2}-16 y^{2} a^{2} x +2 a \,y^{4}+2 a +y \right )^{2}}+\frac {2 a \left (192 a^{3} x^{2} y -96 a^{2} x \,y^{3}+12 a \,y^{5}-32 a^{2} x y +8 a \,y^{3}+1\right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{\left (-128 a^{4} x^{3}+96 y^{2} a^{3} x^{2}-24 a^{2} x \,y^{4}+2 a \,y^{6}+32 a^{3} x^{2}-16 y^{2} a^{2} x +2 a \,y^{4}+2 a +y \right )^{2}} = 0 \end{equation} Putting the above in normal form gives \[ \frac {16384 a^{8} x^{6} b_{2}-24576 a^{7} x^{5} y^{2} b_{2}+15360 a^{6} x^{4} y^{4} b_{2}-5120 a^{5} x^{3} y^{6} b_{2}+960 a^{4} x^{2} y^{8} b_{2}-96 a^{3} x \,y^{10} b_{2}+4 a^{2} y^{12} b_{2}-8192 a^{7} x^{5} b_{2}+10240 a^{6} x^{4} y^{2} b_{2}-5120 a^{5} x^{3} y^{4} b_{2}+1280 a^{4} x^{2} y^{6} b_{2}-160 a^{3} x \,y^{8} b_{2}+8 a^{2} y^{10} b_{2}+1024 a^{6} x^{4} b_{2}-1024 a^{5} x^{3} y^{2} b_{2}+384 a^{4} x^{2} y^{4} b_{2}-64 a^{3} x \,y^{6} b_{2}+4 a^{2} y^{8} b_{2}-512 a^{5} x^{3} a_{2}-512 a^{5} x^{3} b_{2}-256 a^{5} x^{3} b_{3}-768 a^{5} x^{2} y a_{3}+128 a^{4} x^{3} y b_{2}+192 a^{4} x^{2} y^{2} a_{2}+384 a^{4} x^{2} y^{2} b_{2}+576 a^{4} x^{2} y^{2} b_{3}+384 a^{4} x \,y^{3} a_{3}-96 a^{3} x \,y^{4} b_{2}-240 a^{3} x \,y^{4} b_{3}-48 a^{3} y^{5} a_{3}-24 a^{2} x \,y^{5} b_{2}-4 a^{2} y^{6} a_{2}+8 a^{2} y^{6} b_{2}+28 a^{2} y^{6} b_{3}+4 a \,y^{7} b_{2}-768 a^{5} x^{2} a_{1}+384 a^{4} x^{2} y b_{1}+384 a^{4} x \,y^{2} a_{1}-192 a^{3} x \,y^{3} b_{1}-48 a^{3} y^{4} a_{1}+24 a^{2} y^{5} b_{1}+64 a^{4} x^{2} a_{2}+128 a^{4} x^{2} b_{2}+64 a^{4} x^{2} b_{3}+128 a^{4} x y a_{3}-64 a^{3} x \,y^{2} b_{2}-96 a^{3} x \,y^{2} b_{3}-32 a^{3} y^{3} a_{3}-16 a^{2} x \,y^{3} b_{2}-4 a^{2} y^{4} a_{2}+8 a^{2} y^{4} b_{2}+20 a^{2} y^{4} b_{3}+4 a \,y^{5} b_{2}+128 a^{4} x a_{1}-64 a^{3} x y b_{1}-32 a^{3} y^{2} a_{1}+16 a^{2} y^{3} b_{1}-4 a^{2} a_{2}-4 a^{2} a_{3}+4 a^{2} b_{2}+4 a^{2} b_{3}+2 a x b_{2}-2 a y a_{2}+4 a y b_{2}+4 a y b_{3}+y^{2} b_{2}+2 a b_{1}}{\left (128 a^{4} x^{3}-96 y^{2} a^{3} x^{2}+24 a^{2} x \,y^{4}-2 a \,y^{6}-32 a^{3} x^{2}+16 y^{2} a^{2} x -2 a \,y^{4}-2 a -y \right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 16384 a^{8} x^{6} b_{2}-24576 a^{7} x^{5} y^{2} b_{2}+15360 a^{6} x^{4} y^{4} b_{2}-5120 a^{5} x^{3} y^{6} b_{2}+960 a^{4} x^{2} y^{8} b_{2}-96 a^{3} x \,y^{10} b_{2}+4 a^{2} y^{12} b_{2}-8192 a^{7} x^{5} b_{2}+10240 a^{6} x^{4} y^{2} b_{2}-5120 a^{5} x^{3} y^{4} b_{2}+1280 a^{4} x^{2} y^{6} b_{2}-160 a^{3} x \,y^{8} b_{2}+8 a^{2} y^{10} b_{2}+1024 a^{6} x^{4} b_{2}-1024 a^{5} x^{3} y^{2} b_{2}+384 a^{4} x^{2} y^{4} b_{2}-64 a^{3} x \,y^{6} b_{2}+4 a^{2} y^{8} b_{2}-512 a^{5} x^{3} a_{2}-512 a^{5} x^{3} b_{2}-256 a^{5} x^{3} b_{3}-768 a^{5} x^{2} y a_{3}+128 a^{4} x^{3} y b_{2}+192 a^{4} x^{2} y^{2} a_{2}+384 a^{4} x^{2} y^{2} b_{2}+576 a^{4} x^{2} y^{2} b_{3}+384 a^{4} x \,y^{3} a_{3}-96 a^{3} x \,y^{4} b_{2}-240 a^{3} x \,y^{4} b_{3}-48 a^{3} y^{5} a_{3}-24 a^{2} x \,y^{5} b_{2}-4 a^{2} y^{6} a_{2}+8 a^{2} y^{6} b_{2}+28 a^{2} y^{6} b_{3}+4 a \,y^{7} b_{2}-768 a^{5} x^{2} a_{1}+384 a^{4} x^{2} y b_{1}+384 a^{4} x \,y^{2} a_{1}-192 a^{3} x \,y^{3} b_{1}-48 a^{3} y^{4} a_{1}+24 a^{2} y^{5} b_{1}+64 a^{4} x^{2} a_{2}+128 a^{4} x^{2} b_{2}+64 a^{4} x^{2} b_{3}+128 a^{4} x y a_{3}-64 a^{3} x \,y^{2} b_{2}-96 a^{3} x \,y^{2} b_{3}-32 a^{3} y^{3} a_{3}-16 a^{2} x \,y^{3} b_{2}-4 a^{2} y^{4} a_{2}+8 a^{2} y^{4} b_{2}+20 a^{2} y^{4} b_{3}+4 a \,y^{5} b_{2}+128 a^{4} x a_{1}-64 a^{3} x y b_{1}-32 a^{3} y^{2} a_{1}+16 a^{2} y^{3} b_{1}-4 a^{2} a_{2}-4 a^{2} a_{3}+4 a^{2} b_{2}+4 a^{2} b_{3}+2 a x b_{2}-2 a y a_{2}+4 a y b_{2}+4 a y b_{3}+y^{2} b_{2}+2 a b_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 16384 a^{8} b_{2} v_{1}^{6}-24576 a^{7} b_{2} v_{1}^{5} v_{2}^{2}+15360 a^{6} b_{2} v_{1}^{4} v_{2}^{4}-5120 a^{5} b_{2} v_{1}^{3} v_{2}^{6}+960 a^{4} b_{2} v_{1}^{2} v_{2}^{8}-96 a^{3} b_{2} v_{1} v_{2}^{10}+4 a^{2} b_{2} v_{2}^{12}-8192 a^{7} b_{2} v_{1}^{5}+10240 a^{6} b_{2} v_{1}^{4} v_{2}^{2}-5120 a^{5} b_{2} v_{1}^{3} v_{2}^{4}+1280 a^{4} b_{2} v_{1}^{2} v_{2}^{6}-160 a^{3} b_{2} v_{1} v_{2}^{8}+8 a^{2} b_{2} v_{2}^{10}+1024 a^{6} b_{2} v_{1}^{4}-1024 a^{5} b_{2} v_{1}^{3} v_{2}^{2}+384 a^{4} b_{2} v_{1}^{2} v_{2}^{4}-64 a^{3} b_{2} v_{1} v_{2}^{6}+4 a^{2} b_{2} v_{2}^{8}-512 a^{5} a_{2} v_{1}^{3}-768 a^{5} a_{3} v_{1}^{2} v_{2}-512 a^{5} b_{2} v_{1}^{3}-256 a^{5} b_{3} v_{1}^{3}+192 a^{4} a_{2} v_{1}^{2} v_{2}^{2}+384 a^{4} a_{3} v_{1} v_{2}^{3}+128 a^{4} b_{2} v_{1}^{3} v_{2}+384 a^{4} b_{2} v_{1}^{2} v_{2}^{2}+576 a^{4} b_{3} v_{1}^{2} v_{2}^{2}-48 a^{3} a_{3} v_{2}^{5}-96 a^{3} b_{2} v_{1} v_{2}^{4}-240 a^{3} b_{3} v_{1} v_{2}^{4}-4 a^{2} a_{2} v_{2}^{6}-24 a^{2} b_{2} v_{1} v_{2}^{5}+8 a^{2} b_{2} v_{2}^{6}+28 a^{2} b_{3} v_{2}^{6}+4 a b_{2} v_{2}^{7}-768 a^{5} a_{1} v_{1}^{2}+384 a^{4} a_{1} v_{1} v_{2}^{2}+384 a^{4} b_{1} v_{1}^{2} v_{2}-48 a^{3} a_{1} v_{2}^{4}-192 a^{3} b_{1} v_{1} v_{2}^{3}+24 a^{2} b_{1} v_{2}^{5}+64 a^{4} a_{2} v_{1}^{2}+128 a^{4} a_{3} v_{1} v_{2}+128 a^{4} b_{2} v_{1}^{2}+64 a^{4} b_{3} v_{1}^{2}-32 a^{3} a_{3} v_{2}^{3}-64 a^{3} b_{2} v_{1} v_{2}^{2}-96 a^{3} b_{3} v_{1} v_{2}^{2}-4 a^{2} a_{2} v_{2}^{4}-16 a^{2} b_{2} v_{1} v_{2}^{3}+8 a^{2} b_{2} v_{2}^{4}+20 a^{2} b_{3} v_{2}^{4}+4 a b_{2} v_{2}^{5}+128 a^{4} a_{1} v_{1}-32 a^{3} a_{1} v_{2}^{2}-64 a^{3} b_{1} v_{1} v_{2}+16 a^{2} b_{1} v_{2}^{3}-4 a^{2} a_{2}-4 a^{2} a_{3}+4 a^{2} b_{2}+4 a^{2} b_{3}-2 a a_{2} v_{2}+2 a b_{2} v_{1}+4 a b_{2} v_{2}+4 a b_{3} v_{2}+b_{2} v_{2}^{2}+2 a b_{1} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -4 a^{2} a_{2}+4 a^{2} b_{2}+4 a^{2} b_{3}+2 a b_{1}+\left (192 a^{4} a_{2}+384 a^{4} b_{2}+576 a^{4} b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-768 a^{5} a_{3}+384 a^{4} b_{1}\right ) v_{1}^{2} v_{2}-24576 a^{7} b_{2} v_{1}^{5} v_{2}^{2}+15360 a^{6} b_{2} v_{1}^{4} v_{2}^{4}-5120 a^{5} b_{2} v_{1}^{3} v_{2}^{6}+960 a^{4} b_{2} v_{1}^{2} v_{2}^{8}-96 a^{3} b_{2} v_{1} v_{2}^{10}+10240 a^{6} b_{2} v_{1}^{4} v_{2}^{2}-5120 a^{5} b_{2} v_{1}^{3} v_{2}^{4}+1280 a^{4} b_{2} v_{1}^{2} v_{2}^{6}-160 a^{3} b_{2} v_{1} v_{2}^{8}-1024 a^{5} b_{2} v_{1}^{3} v_{2}^{2}+384 a^{4} b_{2} v_{1}^{2} v_{2}^{4}-64 a^{3} b_{2} v_{1} v_{2}^{6}+128 a^{4} b_{2} v_{1}^{3} v_{2}-24 a^{2} b_{2} v_{1} v_{2}^{5}+\left (-96 a^{3} b_{2}-240 a^{3} b_{3}\right ) v_{1} v_{2}^{4}+\left (384 a^{4} a_{3}-192 a^{3} b_{1}-16 a^{2} b_{2}\right ) v_{1} v_{2}^{3}+\left (384 a^{4} a_{1}-64 a^{3} b_{2}-96 a^{3} b_{3}\right ) v_{1} v_{2}^{2}+\left (128 a^{4} a_{3}-64 a^{3} b_{1}\right ) v_{1} v_{2}+\left (-512 a^{5} a_{2}-512 a^{5} b_{2}-256 a^{5} b_{3}\right ) v_{1}^{3}+\left (-768 a^{5} a_{1}+64 a^{4} a_{2}+128 a^{4} b_{2}+64 a^{4} b_{3}\right ) v_{1}^{2}+\left (128 a^{4} a_{1}+2 a b_{2}\right ) v_{1}+\left (-4 a^{2} a_{2}+8 a^{2} b_{2}+28 a^{2} b_{3}\right ) v_{2}^{6}+\left (-48 a^{3} a_{3}+24 a^{2} b_{1}+4 a b_{2}\right ) v_{2}^{5}+\left (-48 a^{3} a_{1}-4 a^{2} a_{2}+8 a^{2} b_{2}+20 a^{2} b_{3}\right ) v_{2}^{4}+\left (-32 a^{3} a_{3}+16 a^{2} b_{1}\right ) v_{2}^{3}+\left (-32 a^{3} a_{1}+b_{2}\right ) v_{2}^{2}+\left (-2 a a_{2}+4 a b_{2}+4 a b_{3}\right ) v_{2}+16384 a^{8} b_{2} v_{1}^{6}+4 a^{2} b_{2} v_{2}^{12}-8192 a^{7} b_{2} v_{1}^{5}+8 a^{2} b_{2} v_{2}^{10}+1024 a^{6} b_{2} v_{1}^{4}+4 a^{2} b_{2} v_{2}^{8}+4 a b_{2} v_{2}^{7}-4 a^{2} a_{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 4 a b_{2}&=0\\ -24 a^{2} b_{2}&=0\\ 4 a^{2} b_{2}&=0\\ 8 a^{2} b_{2}&=0\\ -160 a^{3} b_{2}&=0\\ -96 a^{3} b_{2}&=0\\ -64 a^{3} b_{2}&=0\\ 128 a^{4} b_{2}&=0\\ 384 a^{4} b_{2}&=0\\ 960 a^{4} b_{2}&=0\\ 1280 a^{4} b_{2}&=0\\ -5120 a^{5} b_{2}&=0\\ -1024 a^{5} b_{2}&=0\\ 1024 a^{6} b_{2}&=0\\ 10240 a^{6} b_{2}&=0\\ 15360 a^{6} b_{2}&=0\\ -24576 a^{7} b_{2}&=0\\ -8192 a^{7} b_{2}&=0\\ 16384 a^{8} b_{2}&=0\\ -32 a^{3} a_{1}+b_{2}&=0\\ 128 a^{4} a_{1}+2 a b_{2}&=0\\ -32 a^{3} a_{3}+16 a^{2} b_{1}&=0\\ 128 a^{4} a_{3}-64 a^{3} b_{1}&=0\\ -768 a^{5} a_{3}+384 a^{4} b_{1}&=0\\ -96 a^{3} b_{2}-240 a^{3} b_{3}&=0\\ 384 a^{4} a_{1}-64 a^{3} b_{2}-96 a^{3} b_{3}&=0\\ -2 a a_{2}+4 a b_{2}+4 a b_{3}&=0\\ -4 a^{2} a_{2}+8 a^{2} b_{2}+28 a^{2} b_{3}&=0\\ 192 a^{4} a_{2}+384 a^{4} b_{2}+576 a^{4} b_{3}&=0\\ -512 a^{5} a_{2}-512 a^{5} b_{2}-256 a^{5} b_{3}&=0\\ -48 a^{3} a_{3}+24 a^{2} b_{1}+4 a b_{2}&=0\\ 384 a^{4} a_{3}-192 a^{3} b_{1}-16 a^{2} b_{2}&=0\\ -48 a^{3} a_{1}-4 a^{2} a_{2}+8 a^{2} b_{2}+20 a^{2} b_{3}&=0\\ -768 a^{5} a_{1}+64 a^{4} a_{2}+128 a^{4} b_{2}+64 a^{4} b_{3}&=0\\ -4 a^{2} a_{2}-4 a^{2} a_{3}+4 a^{2} b_{2}+4 a^{2} b_{3}+2 a b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=a_{3}\\ b_{1}&=2 a_{3} a\\ b_{2}&=0\\ 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 &= 2 a \\ \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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime } y^{6} a -24 y^{\prime } y^{4} a^{2} x +96 y^{\prime } y^{2} a^{3} x^{2}-128 x^{3} y^{\prime } a^{4}+2 y^{\prime } y^{4} a -16 y^{\prime } y^{2} a^{2} x +32 y^{\prime } a^{3} x^{2}+y y^{\prime }+2 a y^{\prime }-2 a =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 a}{2 y^{6} a -24 y^{4} a^{2} x +96 y^{2} a^{3} x^{2}-128 a^{4} x^{3}+2 a y^{4}-16 y^{2} a^{2} x +32 a^{3} x^{2}+y+2 a} \end {array} \]
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: 0.047 (sec). Leaf size: 41
dsolve(diff(y(x),x) = -2*a/(-y(x)-2*a-2*a*y(x)^4+16*a^2*x*y(x)^2-32*a^3*x^2-2*a*y(x)^6+24*y(x)^4*a^2*x-96*y(x)^2*a^3*x^2+128*a^4*x^3),y(x), singsol=all)
\[ \frac {y \left (x \right )}{2 a}+\frac {\int _{}^{-4 a x +y \left (x \right )^{2}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}+1}d \textit {\_a}}{8 a^{2}}-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 0.361 (sec). Leaf size: 131
DSolve[y'[x] == (-2*a)/(-2*a - 32*a^3*x^2 + 128*a^4*x^3 - y[x] + 16*a^2*x*y[x]^2 - 96*a^3*x^2*y[x]^2 - 2*a*y[x]^4 + 24*a^2*x*y[x]^4 - 2*a*y[x]^6),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {\text {RootSum}\left [-64 \text {$\#$1}^3 a^3+48 \text {$\#$1}^2 a^2 y(x)^2+16 \text {$\#$1}^2 a^2-12 \text {$\#$1} a y(x)^4-8 \text {$\#$1} a y(x)^2+y(x)^6+y(x)^4+1\&,\frac {\log (x-\text {$\#$1})}{48 \text {$\#$1}^2 a^2-24 \text {$\#$1} a y(x)^2-8 \text {$\#$1} a+3 y(x)^4+2 y(x)^2}\&\right ]}{8 a^2}+\frac {y(x)}{2 a}=c_1,y(x)\right ] \]