2.7.1 Problem 1 (eq 98)

Internal problem ID [18500]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter V. Singular solutions. section 36. Problems at page 99
Problem number : 1 (eq 98)
Date solved : Thursday, March 13, 2025 at 12:08:49 PM
CAS classification : [[_1st_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}4y{y^{\prime }}^3-2x^2{y^{\prime }}^2+4xy{y}^{\prime }+x^3& =16y^2\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& y^{\prime }& =\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}{6y}-\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}\text{(2)}& y^{\prime }& =-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}+\frac{i\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}})}{2}\text{(3)}& y^{\prime }& =-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}-\frac{i\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}})}{2}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Solving for $y^{\prime }$ gives

$$\begin{array}{}\text{(1)}& y^{\prime }& =\frac{x^4+x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}-12x{y}^2+{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{2/3}}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}\end{array}$$

Each of the above ode’s is now solved An ode $y^{\prime }=f(x,y)$ is isobaric if

$$\begin{array}{}\text{(1)}& f(tx,t^{m}y)=t^{m-1}f(x,y)\end{array}$$

Where here

$$\begin{array}{}\text{(2)}& f(x,y)=\frac{x^4+x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}-12x{y}^2+{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{2/3}}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376y^4{x}^3+6912y^6}y)}^{1/3}}\end{array}$$

$m$ is the order of isobaric. Substituting (2) into (1) and solving for $m$ gives

$$m=\frac{3}{2}$$

Since the ode is isobaric of order $m=\frac{3}{2}$, then the substitution

$$\begin{array}{rl}y& =u{x}^m\\ & =u{x}^{3/2}\end{array}$$

Converts the ODE to a separable in $u\left(x\right)$. Performing this substitution gives

$$\frac{3\sqrt{x}u\left(x\right)}{2}+x^{3/2}{u}^{\prime }\left(x\right)=\frac{x^4+x^2{(x^6-45x^{6}u{\left(x\right)}^2+432x^{6}u{\left(x\right)}^4+3\sqrt{3}\sqrt{-2x^9+91x^{9}u{\left(x\right)}^2-1376x^{9}u{\left(x\right)}^4+6912x^{9}u{\left(x\right)}^6}{x}^{3/2}u\left(x\right))}^{1/3}-12x^{4}u{\left(x\right)}^2+{(x^6-45x^{6}u{\left(x\right)}^2+432x^{6}u{\left(x\right)}^4+3\sqrt{3}\sqrt{-2x^9+91x^{9}u{\left(x\right)}^2-1376x^{9}u{\left(x\right)}^4+6912x^{9}u{\left(x\right)}^6}{x}^{3/2}u\left(x\right))}^{2/3}}{6x^{3/2}u\left(x\right){(x^6-45x^{6}u{\left(x\right)}^2+432x^{6}u{\left(x\right)}^4+3\sqrt{3}\sqrt{-2x^9+91x^{9}u{\left(x\right)}^2-1376x^{9}u{\left(x\right)}^4+6912x^{9}u{\left(x\right)}^6}{x}^{3/2}u\left(x\right))}^{1/3}}$$

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }\left(x\right)=\frac{(-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}u{\left(x\right)}^2-123^{2/3}u{\left(x\right)}^2+{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3})3^{2/3}}{18{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}u\left(x\right)x}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }\left(x\right)& =\frac{(-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}u{\left(x\right)}^2-123^{2/3}u{\left(x\right)}^2+{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3})3^{2/3}}{18{\left(\sqrt{3}(432\sqrt{3}u{\left(x\right)}^4-45\sqrt{3}u{\left(x\right)}^2+9u\left(x\right)\sqrt{(27u{\left(x\right)}^2-2){(4u\left(x\right)-1)}^2{(4u\left(x\right)+1)}^2}+\sqrt{3})\right)}^{1/3}u\left(x\right)x}\\ & =f(x)g(u)\end{array}$$

Where

$$\begin{array}{rl}f(x)& =\frac{3^{2/3}}{18x}\\ g(u)& =\frac{-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}{u}^2-123^{2/3}{u}^2+{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3}}{{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}u}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(x)dx\\ \int \frac{{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}u}{-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}{u}^2-123^{2/3}{u}^2+{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}{u}^4-45\sqrt{3}{u}^2+9u\sqrt{(27u^2-2){(4u-1)}^2{(4u+1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3}}du& =\int \frac{3^{2/3}}{18x}dx\end{array}$$

$${\int }_{}^{u\left(x\right)}\frac{{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}\tau }{-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}{\tau }^2-123^{2/3}{\tau }^2+{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3}}d\tau =3^{2/3}\ln \left(x^{1/18}\right)+c_1$$

Converting ${\int }_{}^{u\left(x\right)}\frac{{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}\tau }{-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}{\tau }^2-123^{2/3}{\tau }^2+{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3}}d\tau =\frac{3^{2/3}\ln \left(x\right)}{18}+c_1$ back to $y$ gives

$$\begin{array}{r}{\int }_{}^{\frac{y}{x^{3/2}}}\frac{{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}\tau }{-93^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}{\tau }^2-123^{2/3}{\tau }^2+{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{2/3}+3^{1/3}{\left(\sqrt{3}(432\sqrt{3}{\tau }^4-45\sqrt{3}{\tau }^2+9\tau \sqrt{(27{\tau }^2-2){(4\tau -1)}^2{(4\tau +1)}^2}+\sqrt{3})\right)}^{1/3}+3^{2/3}}d\tau =\frac{3^{2/3}\ln \left(x\right)}{18}+c_1\end{array}$$

Solving Eq. (2)

Writing the ode as

$$\begin{array}{rl}{y}^{\prime }& =\frac{-i\sqrt{3}{x}^4+12i\sqrt{3}x{y}^2+i\sqrt{3}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}-x^4+2x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}+12x{y}^2-{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}}\\ y^{\prime }& =\omega (x,y)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

$$\begin{array}{}\text{(A)}& {\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta =0\end{array}$$

To determine $\xi ,\eta$ then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

$$\begin{array}{}\text{(1E)}& \xi & =x{a}_2+y{a}_3+a_1\text{(2E)}& \eta & =x{b}_2+y{b}_3+b_1\end{array}$$

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

$$\begin{array}{}\text{(5E)}& \text{Expression too large to display}\end{array}$$

Putting the above in normal form gives

$$\text{Expression too large to display}$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& \text{Expression too large to display}\end{array}$$

Simplifying the above gives

$$\begin{array}{}\text{(6E)}& \text{Expression too large to display}\end{array}$$

Since the PDE has radicals, simplifying gives

$$\text{Expression too large to display}$$

Looking at the above PDE shows the following are all the terms with $\{x,y\}$ in them.

$$\{x,y,\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2},{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{1/3},{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{2/3}\}$$

The following substitution is now made to be able to collect on all terms with $\{x,y\}$ in them

$$\{x=v_1,y=v_2,\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}=v_3,{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{1/3}=v_4,{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{2/3}=v_5\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& \text{Expression too large to display}\end{array}$$

Collecting the above on the terms $v_i$ introduced, and these are

$$\{v_1,v_2,v_3,v_4,v_5\}$$

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& \text{Expression too large to display}\end{array}$$

Setting each coefficients in (8E) to zero gives the following equations to solve

$$\begin{array}{rl}-20736a_1& =0\\ -48a_1& =0\\ 2160a_1& =0\\ -7344a_3& =0\\ -12a_3& =0\\ 588a_3& =0\\ 20736a_3& =0\\ -1080b_1& =0\\ 24b_1& =0\\ 10368b_1& =0\\ -936b_2& =0\\ 24b_2& =0\\ 3888b_2& =0\\ 62208b_2& =0\\ -995328\sqrt{3}{a}_1& =0\\ -13104\sqrt{3}{a}_1& =0\\ 288\sqrt{3}{a}_1& =0\\ 198144\sqrt{3}{a}_1& =0\\ -446976\sqrt{3}{a}_3& =0\\ -3564\sqrt{3}{a}_3& =0\\ 72\sqrt{3}{a}_3& =0\\ 62640\sqrt{3}{a}_3& =0\\ 995328\sqrt{3}{a}_3& =0\\ -99072\sqrt{3}{b}_1& =0\\ -144\sqrt{3}{b}_1& =0\\ 6552\sqrt{3}{b}_1& =0\\ 497664\sqrt{3}{b}_1& =0\\ -96768\sqrt{3}{b}_2& =0\\ -59760\sqrt{3}{b}_2& =0\\ -144\sqrt{3}{b}_2& =0\\ 5688\sqrt{3}{b}_2& =0\\ 2985984\sqrt{3}{b}_2& =0\\ -31104a_2+20736b_3& =0\\ -72a_2+48b_3& =0\\ 3240a_2-2160b_3& =0\\ -1492992\sqrt{3}{a}_2+995328\sqrt{3}{b}_3& =0\\ -19656\sqrt{3}{a}_2+13104\sqrt{3}{b}_3& =0\\ 432\sqrt{3}{a}_2-288\sqrt{3}{b}_3& =0\\ 297216\sqrt{3}{a}_2-198144\sqrt{3}{b}_3& =0\\ -31705344i{a}_3-10568448\sqrt{3}{a}_3& =0\\ -8957952i{a}_1-2985984\sqrt{3}{a}_1& =0\\ -1741824i{b}_1-580608\sqrt{3}{b}_1& =0\\ -1741824i{b}_2-580608\sqrt{3}{b}_2& =0\\ -520128i{a}_1-173376\sqrt{3}{a}_1& =0\\ -288720i{a}_3-96240\sqrt{3}{a}_3& =0\\ -248832i{b}_1+82944\sqrt{3}{b}_1& =0\\ -248832i{b}_2+82944\sqrt{3}{b}_2& =0\\ -174528i{a}_3+58176\sqrt{3}{a}_3& =0\\ -14652i{b}_1-4884\sqrt{3}{b}_1& =0\\ -14652i{b}_2-4884\sqrt{3}{b}_2& =0\\ -9828i{a}_1+3276\sqrt{3}{a}_1& =0\\ -2952i{a}_3+984\sqrt{3}{a}_3& =0\\ -540i{a}_1-180\sqrt{3}{a}_1& =0\\ -144i{a}_3-48\sqrt{3}{a}_3& =0\\ -144i{b}_1+48\sqrt{3}{b}_1& =0\\ -144i{b}_2+48\sqrt{3}{b}_2& =0\\ 72i{a}_3-24\sqrt{3}{a}_3& =0\\ 288i{b}_1+96\sqrt{3}{b}_1& =0\\ 288i{b}_2+96\sqrt{3}{b}_2& =0\\ 324i{a}_1-108\sqrt{3}{a}_1& =0\\ 3276i{b}_1-1092\sqrt{3}{b}_1& =0\\ 3276i{b}_2-1092\sqrt{3}{b}_2& =0\\ 10116i{a}_3+3372\sqrt{3}{a}_3& =0\\ 28188i{a}_1+9396\sqrt{3}{a}_1& =0\\ 39708i{a}_3-13236\sqrt{3}{a}_3& =0\\ 74304i{a}_1-24768\sqrt{3}{a}_1& =0\\ 257904i{b}_1+85968\sqrt{3}{b}_1& =0\\ 257904i{b}_2+85968\sqrt{3}{b}_2& =0\\ 2985984i{b}_1+995328\sqrt{3}{b}_1& =0\\ 2985984i{b}_2+995328\sqrt{3}{b}_2& =0\\ 3877632i{a}_1+1292544\sqrt{3}{a}_1& =0\\ 4216320i{a}_3+1405440\sqrt{3}{a}_3& =0\\ 98537472i{a}_3+32845824\sqrt{3}{a}_3& =0\\ -152064i\sqrt{3}{a}_3-152064a_3& =0\\ -62208i\sqrt{3}{a}_1-62208a_1& =0\\ -9936i\sqrt{3}{b}_1-9936b_1& =0\\ -9936i\sqrt{3}{b}_2-9936b_2& =0\\ -1728i\sqrt{3}{b}_1+1728b_1& =0\\ -1728i\sqrt{3}{b}_2+1728b_2& =0\\ -1476i\sqrt{3}{a}_1-1476a_1& =0\\ -1476i\sqrt{3}{a}_3+1476a_3& =0\\ -492i\sqrt{3}{a}_3-492a_3& =0\\ -180i\sqrt{3}{b}_1+180b_1& =0\\ -180i\sqrt{3}{b}_2+180b_2& =0\\ -24i\sqrt{3}{a}_1+24a_1& =0\\ -12i\sqrt{3}{b}_1-12b_1& =0\\ -12i\sqrt{3}{b}_2-12b_2& =0\\ -6i\sqrt{3}{a}_3+6a_3& =0\\ 6i\sqrt{3}{a}_3+6a_3& =0\\ 12i\sqrt{3}{b}_1-12b_1& =0\\ 12i\sqrt{3}{b}_2-12b_2& =0\\ 24i\sqrt{3}{a}_1+24a_1& =0\\ 204i\sqrt{3}{a}_3-204a_3& =0\\ 540i\sqrt{3}{a}_1-540a_1& =0\\ 756i\sqrt{3}{b}_1+756b_1& =0\\ 756i\sqrt{3}{b}_2+756b_2& =0\\ 13104i\sqrt{3}{a}_3+13104a_3& =0\\ 20736i\sqrt{3}{a}_1+20736a_1& =0\\ 20736i\sqrt{3}{b}_1+20736b_1& =0\\ 20736i\sqrt{3}{b}_2+20736b_2& =0\\ 684288i\sqrt{3}{a}_3+684288a_3& =0\\ -17915904i{a}_2+11943936i{b}_3-5971968\sqrt{3}{a}_2+3981312\sqrt{3}{b}_3& =0\\ -786672i{a}_2+524448i{b}_3-262224\sqrt{3}{a}_2+174816\sqrt{3}{b}_3& =0\\ -746496i{a}_2+497664i{b}_3+248832\sqrt{3}{a}_2-165888\sqrt{3}{b}_3& =0\\ -19656i{a}_2+13104i{b}_3+6552\sqrt{3}{a}_2-4368\sqrt{3}{b}_3& =0\\ -756i{a}_2+504i{b}_3-252\sqrt{3}{a}_2+168\sqrt{3}{b}_3& =0\\ 540i{a}_2-360i{b}_3-180\sqrt{3}{a}_2+120\sqrt{3}{b}_3& =0\\ 40608i{a}_2-27072i{b}_3+13536\sqrt{3}{a}_2-9024\sqrt{3}{b}_3& =0\\ 222912i{a}_2-148608i{b}_3-74304\sqrt{3}{a}_2+49536\sqrt{3}{b}_3& =0\\ 6407424i{a}_2-4271616i{b}_3+2135808\sqrt{3}{a}_2-1423872\sqrt{3}{b}_3& =0\\ -124416i\sqrt{3}{a}_2+82944i\sqrt{3}{b}_3-124416a_2+82944b_3& =0\\ -5184i\sqrt{3}{a}_2+3456i\sqrt{3}{b}_3+5184a_2-3456b_3& =0\\ -2160i\sqrt{3}{a}_2+1440i\sqrt{3}{b}_3-2160a_2+1440b_3& =0\\ -36i\sqrt{3}{a}_2+24i\sqrt{3}{b}_3+36a_2-24b_3& =0\\ 36i\sqrt{3}{a}_2-24i\sqrt{3}{b}_3+36a_2-24b_3& =0\\ 1080i\sqrt{3}{a}_2-720i\sqrt{3}{b}_3-1080a_2+720b_3& =0\\ 32400i\sqrt{3}{a}_2-21600i\sqrt{3}{b}_3+32400a_2-21600b_3& =0\end{array}$$

Solving the above equations for the unknowns gives

$$\begin{array}{rl}{a}_1& =0\\ a_2& =\frac{2b_3}{3}\\ a_3& =0\\ b_1& =0\\ b_2& =0\\ b_3& =b_3\end{array}$$

Substituting the above solution in the anstaz (1E,2E) (using $1$ as arbitrary value for any unknown in the RHS) gives

$$\begin{array}{rl}\xi & =\frac{2x}{3}\\ \eta & =y\end{array}$$

The next step is to determine the canonical coordinates $R,S$. The canonical coordinates map $(x,y)\to (R,S)$ where $(R,S)$ 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

$$\begin{array}{}\text{(1)}& \frac{dx}{\xi }& =\frac{dy}{\eta }=dS\end{array}$$

The above comes from the requirements that $(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y})S(x,y)=1$. 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 $S(R)$. Therefore

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{\eta }{\xi }\\ & =\frac{y}{\frac{2x}{3}}\\ & =\frac{3y}{2x}\end{array}$$

This is easily solved to give

$$\begin{array}{r}y=c_1{x}^{3/2}\end{array}$$

Where now the coordinate $R$ is taken as the constant of integration. Hence

$$\begin{array}{rl}R& =\frac{y}{x^{3/2}}\end{array}$$

And $S$ is found from

$$\begin{array}{rl}dS& =\frac{dx}{\xi }\\ & =\frac{dx}{\frac{2x}{3}}\end{array}$$

Integrating gives

$$\begin{array}{rl}S& =\int \frac{dx}{T}\\ & =\frac{3\ln \left(x\right)}{2}\end{array}$$

Where the constant of integration is set to zero as we just need one solution. Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_x+\omega (x,y)S_y}{R_x+\omega (x,y)R_y}\end{array}$$

Where in the above $R_x,R_y,S_x,S_y$ are all partial derivatives and $\omega (x,y)$ is the right hand side of the original ode given by

$$\begin{array}{rl}\omega (x,y)& =\frac{-i\sqrt{3}{x}^4+12i\sqrt{3}x{y}^2+i\sqrt{3}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}-x^4+2x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}+12x{y}^2-{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}}\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_x& =-\frac{3y}{2x^{5/2}}\\ R_y& =\frac{1}{x^{3/2}}\\ S_x& =\frac{3}{2x}\\ S_y& =0\end{array}$$

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =-\frac{18x^{3/2}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{1/3}y}{(-i\sqrt{3}x+x){(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{2/3}+(-2x^3+18y^2){(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{1/3}+(x^3-12y^2)x^2(1+i\sqrt{3})}\end{array}$$

We now need to express the RHS as function of $R$ only. This is done by solving for $x,y$ in terms of $R,S$ from the result obtained earlier and simplifying. This gives

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{(i\sqrt{3}-1){((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}+(-18R^2+2){((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}+12(1+i\sqrt{3})(R^2-\frac{1}{12})}\end{array}$$

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates $R,S$.

Since the ode has the form $\frac{d}{dR}S\left(R\right)=f(R)$, then we only need to integrate $f(R)$.

$$\begin{array}{rl}\int dS& =\int \frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}\sqrt{3}+12i\sqrt{3}{R}^2-18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-i\sqrt{3}+12R^2+2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}-1}dR\\ S\left(R\right)& =\int \frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}\sqrt{3}+12i\sqrt{3}{R}^2-18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-i\sqrt{3}+12R^2+2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}-1}dR+c_3\end{array}$$

$$\begin{array}{rl}S\left(R\right)& =\int \frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}\sqrt{3}+12i\sqrt{3}{R}^2-18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-i\sqrt{3}+12R^2+2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}-1}dR+c_3\end{array}$$

To complete the solution, we just need to transform the above back to $x,y$ coordinates. This results in

$$\begin{array}{r}\frac{3\ln \left(x\right)}{2}={\int }_{}^{\frac{y}{x^{3/2}}}\frac{18\mathit{\text{\_a}}{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}}{i{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{2/3}\sqrt{3}+12i\sqrt{3}{\mathit{\text{\_a}}}^2-18{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}{\mathit{\text{\_a}}}^2-{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{2/3}-i\sqrt{3}+12{\mathit{\text{\_a}}}^2+2{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}-1}d\mathit{\text{\_a}}+c_3\end{array}$$

Solving Eq. (3)

Writing the ode as

$$\begin{array}{rl}{y}^{\prime }& =-\frac{-i\sqrt{3}{x}^4+12i\sqrt{3}x{y}^2+i\sqrt{3}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}+x^4-2x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}-12x{y}^2+{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}}\\ y^{\prime }& =\omega (x,y)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

$$\begin{array}{}\text{(A)}& {\eta }_x+\omega ({\eta }_y-{\xi }_x)-{\omega }^2{\xi }_y-{\omega }_x\xi -{\omega }_y\eta =0\end{array}$$

To determine $\xi ,\eta$ then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

$$\begin{array}{}\text{(1E)}& \xi & =x{a}_2+y{a}_3+a_1\text{(2E)}& \eta & =x{b}_2+y{b}_3+b_1\end{array}$$

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

$$\begin{array}{}\text{(5E)}& \text{Expression too large to display}\end{array}$$

Putting the above in normal form gives

$$\text{Expression too large to display}$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& \text{Expression too large to display}\end{array}$$

Simplifying the above gives

$$\begin{array}{}\text{(6E)}& \text{Expression too large to display}\end{array}$$

Since the PDE has radicals, simplifying gives

$$\text{Expression too large to display}$$

Looking at the above PDE shows the following are all the terms with $\{x,y\}$ in them.

$$\{x,y,\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2},{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{1/3},{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{2/3}\}$$

The following substitution is now made to be able to collect on all terms with $\{x,y\}$ in them

$$\{x=v_1,y=v_2,\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}=v_3,{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{1/3}=v_4,{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-(2x^3-27y^2){(x^3-16y^2)}^2}y)}^{2/3}=v_5\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& \text{Expression too large to display}\end{array}$$

Collecting the above on the terms $v_i$ introduced, and these are

$$\{v_1,v_2,v_3,v_4,v_5\}$$

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& \text{Expression too large to display}\end{array}$$

Setting each coefficients in (8E) to zero gives the following equations to solve

$$\begin{array}{rl}-20736a_1& =0\\ -48a_1& =0\\ 2160a_1& =0\\ -7344a_3& =0\\ -12a_3& =0\\ 588a_3& =0\\ 20736a_3& =0\\ -1080b_1& =0\\ 24b_1& =0\\ 10368b_1& =0\\ -936b_2& =0\\ 24b_2& =0\\ 3888b_2& =0\\ 62208b_2& =0\\ -995328\sqrt{3}{a}_1& =0\\ -13104\sqrt{3}{a}_1& =0\\ 288\sqrt{3}{a}_1& =0\\ 198144\sqrt{3}{a}_1& =0\\ -446976\sqrt{3}{a}_3& =0\\ -3564\sqrt{3}{a}_3& =0\\ 72\sqrt{3}{a}_3& =0\\ 62640\sqrt{3}{a}_3& =0\\ 995328\sqrt{3}{a}_3& =0\\ -99072\sqrt{3}{b}_1& =0\\ -144\sqrt{3}{b}_1& =0\\ 6552\sqrt{3}{b}_1& =0\\ 497664\sqrt{3}{b}_1& =0\\ -96768\sqrt{3}{b}_2& =0\\ -59760\sqrt{3}{b}_2& =0\\ -144\sqrt{3}{b}_2& =0\\ 5688\sqrt{3}{b}_2& =0\\ 2985984\sqrt{3}{b}_2& =0\\ -31104a_2+20736b_3& =0\\ -72a_2+48b_3& =0\\ 3240a_2-2160b_3& =0\\ -2985984\sqrt{3}{a}_1+8957952i{a}_1& =0\\ -173376\sqrt{3}{a}_1+520128i{a}_1& =0\\ -24768\sqrt{3}{a}_1-74304i{a}_1& =0\\ -180\sqrt{3}{a}_1+540i{a}_1& =0\\ -108\sqrt{3}{a}_1-324i{a}_1& =0\\ 3276\sqrt{3}{a}_1+9828i{a}_1& =0\\ 9396\sqrt{3}{a}_1-28188i{a}_1& =0\\ 1292544\sqrt{3}{a}_1-3877632i{a}_1& =0\\ -1492992\sqrt{3}{a}_2+995328\sqrt{3}{b}_3& =0\\ -19656\sqrt{3}{a}_2+13104\sqrt{3}{b}_3& =0\\ 432\sqrt{3}{a}_2-288\sqrt{3}{b}_3& =0\\ 297216\sqrt{3}{a}_2-198144\sqrt{3}{b}_3& =0\\ -10568448\sqrt{3}{a}_3+31705344i{a}_3& =0\\ -96240\sqrt{3}{a}_3+288720i{a}_3& =0\\ -13236\sqrt{3}{a}_3-39708i{a}_3& =0\\ -48\sqrt{3}{a}_3+144i{a}_3& =0\\ -24\sqrt{3}{a}_3-72i{a}_3& =0\\ 984\sqrt{3}{a}_3+2952i{a}_3& =0\\ 3372\sqrt{3}{a}_3-10116i{a}_3& =0\\ 58176\sqrt{3}{a}_3+174528i{a}_3& =0\\ 1405440\sqrt{3}{a}_3-4216320i{a}_3& =0\\ 32845824\sqrt{3}{a}_3-98537472i{a}_3& =0\\ -580608\sqrt{3}{b}_1+1741824i{b}_1& =0\\ -4884\sqrt{3}{b}_1+14652i{b}_1& =0\\ -1092\sqrt{3}{b}_1-3276i{b}_1& =0\\ 48\sqrt{3}{b}_1+144i{b}_1& =0\\ 96\sqrt{3}{b}_1-288i{b}_1& =0\\ 82944\sqrt{3}{b}_1+248832i{b}_1& =0\\ 85968\sqrt{3}{b}_1-257904i{b}_1& =0\\ 995328\sqrt{3}{b}_1-2985984i{b}_1& =0\\ -580608\sqrt{3}{b}_2+1741824i{b}_2& =0\\ -4884\sqrt{3}{b}_2+14652i{b}_2& =0\\ -1092\sqrt{3}{b}_2-3276i{b}_2& =0\\ 48\sqrt{3}{b}_2+144i{b}_2& =0\\ 96\sqrt{3}{b}_2-288i{b}_2& =0\\ 82944\sqrt{3}{b}_2+248832i{b}_2& =0\\ 85968\sqrt{3}{b}_2-257904i{b}_2& =0\\ 995328\sqrt{3}{b}_2-2985984i{b}_2& =0\\ -684288i\sqrt{3}{a}_3+684288a_3& =0\\ -20736i\sqrt{3}{a}_1+20736a_1& =0\\ -20736i\sqrt{3}{b}_1+20736b_1& =0\\ -20736i\sqrt{3}{b}_2+20736b_2& =0\\ -13104i\sqrt{3}{a}_3+13104a_3& =0\\ -756i\sqrt{3}{b}_1+756b_1& =0\\ -756i\sqrt{3}{b}_2+756b_2& =0\\ -540i\sqrt{3}{a}_1-540a_1& =0\\ -204i\sqrt{3}{a}_3-204a_3& =0\\ -24i\sqrt{3}{a}_1+24a_1& =0\\ -12i\sqrt{3}{b}_1-12b_1& =0\\ -12i\sqrt{3}{b}_2-12b_2& =0\\ -6i\sqrt{3}{a}_3+6a_3& =0\\ 6i\sqrt{3}{a}_3+6a_3& =0\\ 12i\sqrt{3}{b}_1-12b_1& =0\\ 12i\sqrt{3}{b}_2-12b_2& =0\\ 24i\sqrt{3}{a}_1+24a_1& =0\\ 180i\sqrt{3}{b}_1+180b_1& =0\\ 180i\sqrt{3}{b}_2+180b_2& =0\\ 492i\sqrt{3}{a}_3-492a_3& =0\\ 1476i\sqrt{3}{a}_1-1476a_1& =0\\ 1476i\sqrt{3}{a}_3+1476a_3& =0\\ 1728i\sqrt{3}{b}_1+1728b_1& =0\\ 1728i\sqrt{3}{b}_2+1728b_2& =0\\ 9936i\sqrt{3}{b}_1-9936b_1& =0\\ 9936i\sqrt{3}{b}_2-9936b_2& =0\\ 62208i\sqrt{3}{a}_1-62208a_1& =0\\ 152064i\sqrt{3}{a}_3-152064a_3& =0\\ -5971968\sqrt{3}{a}_2+3981312\sqrt{3}{b}_3+17915904i{a}_2-11943936i{b}_3& =0\\ -262224\sqrt{3}{a}_2+174816\sqrt{3}{b}_3+786672i{a}_2-524448i{b}_3& =0\\ -74304\sqrt{3}{a}_2+49536\sqrt{3}{b}_3-222912i{a}_2+148608i{b}_3& =0\\ -252\sqrt{3}{a}_2+168\sqrt{3}{b}_3+756i{a}_2-504i{b}_3& =0\\ -180\sqrt{3}{a}_2+120\sqrt{3}{b}_3-540i{a}_2+360i{b}_3& =0\\ 6552\sqrt{3}{a}_2-4368\sqrt{3}{b}_3+19656i{a}_2-13104i{b}_3& =0\\ 13536\sqrt{3}{a}_2-9024\sqrt{3}{b}_3-40608i{a}_2+27072i{b}_3& =0\\ 248832\sqrt{3}{a}_2-165888\sqrt{3}{b}_3+746496i{a}_2-497664i{b}_3& =0\\ 2135808\sqrt{3}{a}_2-1423872\sqrt{3}{b}_3-6407424i{a}_2+4271616i{b}_3& =0\\ -32400i\sqrt{3}{a}_2+21600i\sqrt{3}{b}_3+32400a_2-21600b_3& =0\\ -1080i\sqrt{3}{a}_2+720i\sqrt{3}{b}_3-1080a_2+720b_3& =0\\ -36i\sqrt{3}{a}_2+24i\sqrt{3}{b}_3+36a_2-24b_3& =0\\ 36i\sqrt{3}{a}_2-24i\sqrt{3}{b}_3+36a_2-24b_3& =0\\ 2160i\sqrt{3}{a}_2-1440i\sqrt{3}{b}_3-2160a_2+1440b_3& =0\\ 5184i\sqrt{3}{a}_2-3456i\sqrt{3}{b}_3+5184a_2-3456b_3& =0\\ 124416i\sqrt{3}{a}_2-82944i\sqrt{3}{b}_3-124416a_2+82944b_3& =0\end{array}$$

Solving the above equations for the unknowns gives

$$\begin{array}{rl}{a}_1& =0\\ a_2& =\frac{2b_3}{3}\\ a_3& =0\\ b_1& =0\\ b_2& =0\\ b_3& =b_3\end{array}$$

Substituting the above solution in the anstaz (1E,2E) (using $1$ as arbitrary value for any unknown in the RHS) gives

$$\begin{array}{rl}\xi & =\frac{2x}{3}\\ \eta & =y\end{array}$$

The next step is to determine the canonical coordinates $R,S$. The canonical coordinates map $(x,y)\to (R,S)$ where $(R,S)$ 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

$$\begin{array}{}\text{(1)}& \frac{dx}{\xi }& =\frac{dy}{\eta }=dS\end{array}$$

The above comes from the requirements that $(\xi \frac{\partial }{\partial x}+\eta \frac{\partial }{\partial y})S(x,y)=1$. 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 $S(R)$. Therefore

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{\eta }{\xi }\\ & =\frac{y}{\frac{2x}{3}}\\ & =\frac{3y}{2x}\end{array}$$

This is easily solved to give

$$\begin{array}{r}y=c_1{x}^{3/2}\end{array}$$

Where now the coordinate $R$ is taken as the constant of integration. Hence

$$\begin{array}{rl}R& =\frac{y}{x^{3/2}}\end{array}$$

And $S$ is found from

$$\begin{array}{rl}dS& =\frac{dx}{\xi }\\ & =\frac{dx}{\frac{2x}{3}}\end{array}$$

Integrating gives

$$\begin{array}{rl}S& =\int \frac{dx}{T}\\ & =\frac{3\ln \left(x\right)}{2}\end{array}$$

Where the constant of integration is set to zero as we just need one solution. Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_x+\omega (x,y)S_y}{R_x+\omega (x,y)R_y}\end{array}$$

Where in the above $R_x,R_y,S_x,S_y$ are all partial derivatives and $\omega (x,y)$ is the right hand side of the original ode given by

$$\begin{array}{rl}\omega (x,y)& =-\frac{-i\sqrt{3}{x}^4+12i\sqrt{3}x{y}^2+i\sqrt{3}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}+x^4-2x^2{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}-12x{y}^2+{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{2/3}}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91x^6{y}^2-1376x^3{y}^4+6912y^6}y)}^{1/3}}\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_x& =-\frac{3y}{2x^{5/2}}\\ R_y& =\frac{1}{x^{3/2}}\\ S_x& =\frac{3}{2x}\\ S_y& =0\end{array}$$

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =\frac{18x^{3/2}{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{1/3}y}{x(-1-i\sqrt{3}){(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{2/3}+2(x^3-9y^2){(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2{(x^3-16y^2)}^2(x^3-\frac{27y^2}{2})}y)}^{1/3}+(x^3-12y^2)x^2(i\sqrt{3}-1)}\end{array}$$

We now need to express the RHS as function of $R$ only. This is done by solving for $x,y$ in terms of $R,S$ from the result obtained earlier and simplifying. This gives

$$\begin{array}{rl}\frac{dS}{dR}& =-\frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{(1+i\sqrt{3}){((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}+(18R^2-2){((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}+12(i\sqrt{3}-1)(R^2-\frac{1}{12})}\end{array}$$

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates $R,S$.

Since the ode has the form $\frac{d}{dR}S\left(R\right)=f(R)$, then we only need to integrate $f(R)$.

$$\begin{array}{rl}\int dS& =\int -\frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i\sqrt{3}{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}+12i\sqrt{3}{R}^2+18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-i\sqrt{3}+{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-12R^2-2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}+1}dR\\ S\left(R\right)& =\int -\frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i\sqrt{3}{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}+12i\sqrt{3}{R}^2+18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-i\sqrt{3}+{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-12R^2-2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}+1}dR+c_5\end{array}$$

$$\begin{array}{rl}S\left(R\right)& =\int -\frac{18R{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}}{i\sqrt{3}{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}+12i\sqrt{3}{R}^2+18{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}{R}^2-i\sqrt{3}+{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{2/3}-12R^2-2{((48R^3-3R)\sqrt{3}\sqrt{27R^2-2}+432R^4-45R^2+1)}^{1/3}+1}dR+c_5\end{array}$$

To complete the solution, we just need to transform the above back to $x,y$ coordinates. This results in

$$\begin{array}{r}\frac{3\ln \left(x\right)}{2}={\int }_{}^{\frac{y}{x^{3/2}}}-\frac{18\mathit{\text{\_a}}{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}}{i{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{2/3}\sqrt{3}+12i\sqrt{3}{\mathit{\text{\_a}}}^2+18{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}{\mathit{\text{\_a}}}^2-i\sqrt{3}+{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{2/3}-12{\mathit{\text{\_a}}}^2-2{((48{\mathit{\text{\_a}}}^3-3\mathit{\text{\_a}})\sqrt{3}\sqrt{27{\mathit{\text{\_a}}}^2-2}+432{\mathit{\text{\_a}}}^4-45{\mathit{\text{\_a}}}^2+1)}^{1/3}+1}d\mathit{\text{\_a}}+c_5\end{array}$$

✗ Maple

ode:=4*y(x)*diff(y(x),x)^3-2*x^2*diff(y(x),x)^2+4*x*y(x)*diff(y(x),x)+x^3 = 16*y(x)^2; 
dsolve(ode,y(x), singsol=all);
 

$$\text{No solution found}$$

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 4y{y^{\prime }}^3-2{y^{\prime }}^2{x}^2+4xy{y}^{\prime }+x^3=16y^2\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & y^{\prime }\\ \bullet & & \text{Solve for the highest derivative}\\ & & [y^{\prime }=\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}-\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y},y^{\prime }=-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}-\frac{\mathrm{I}\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}})}{2},y^{\prime }=-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}+\frac{\mathrm{I}\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}})}{2}]\\ \bullet & & \text{Solve the equation}{y}^{\prime }=\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}-\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}\\ \bullet & & \text{Solve the equation}{y}^{\prime }=-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}-\frac{\mathrm{I}\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}})}{2}\\ \bullet & & \text{Solve the equation}{y}^{\prime }=-\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{12y}+\frac{x(-x^3+12y^2)}{12y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}+\frac{x^2}{6y}+\frac{\mathrm{I}\sqrt{3}(\frac{{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}}{6y}+\frac{x(-x^3+12y^2)}{6y{(x^6-45x^3{y}^2+432y^4+3\sqrt{3}\sqrt{-2x^9+91y^2{x}^6-1376x^3{y}^4+6912y^6}y)}^{1/3}})}{2}\\ \bullet & & \text{Set of solutions}\\ & & \{\mathit{workingODE},\mathit{workingODE},\mathit{workingODE}\}\end{array}$$

✓ Mathematica. Time used: 50.04 (sec). Leaf size: 49162

ode=4*y[x]*D[y[x],x]^3-2*x^2*D[y[x],x]^2+4*x*y[x]*D[y[x],x]+x^3==16*y[x]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Too large to display

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**3 - 2*x**2*Derivative(y(x), x)**2 + 4*x*y(x)*Derivative(y(x), x) - 16*y(x)**2 + 4*y(x)*Derivative(y(x), x)**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out