2.1.51 Problem 51

Internal problem ID [9122]
Book : Second order enumerated odes
Section : section 1
Problem number : 51
Date solved : Sunday, March 30, 2025 at 02:08:06 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order missing x ode

Time used: 92.645 (sec)

Solve

$$\begin{array}{rl}y{y^{\prime \prime }}^3+y^3{y}^{\prime }& =0\end{array}$$

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable $y$ an independent variable. Using

$$\begin{array}{rl}{y}^{\prime }& =p\end{array}$$

Then

$$\begin{array}{rl}{y}^{″}& =\frac{dp}{dx}\\ & =\frac{dp}{dy}\frac{dy}{dx}\\ & =p\frac{dp}{dy}\end{array}$$

Hence the ode becomes

$$\begin{array}{r}yp{\left(y\right)}^3{\left(\frac{d}{dy}p\left(y\right)\right)}^3+y^{3}p\left(y\right)=0\end{array}$$

Which is now solved as first order ode for $p(y)$.

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& p^{\prime }& =\frac{{(-y^{2}p)}^{1/3}}{p}\text{(2)}& p^{\prime }& =-\frac{{(-y^{2}p)}^{1/3}}{2p}-\frac{i\sqrt{3}{(-y^{2}p)}^{1/3}}{2p}\text{(3)}& p^{\prime }& =-\frac{{(-y^{2}p)}^{1/3}}{2p}+\frac{i\sqrt{3}{(-y^{2}p)}^{1/3}}{2p}\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

In canonical form, the ODE is

$$\begin{array}{rl}{p}^{\prime }& =F(y,p)\\ \text{(1)}& & =\frac{{(-y^{2}p)}^{1/3}}{p}\end{array}$$

An ode of the form $p^{\prime }=\frac{M(y,p)}{N(y,p)}$ is called homogeneous if the functions $M(y,p)$ and $N(y,p)$ are both homogeneous functions and of the same order. Recall that a function $f(y,p)$ is homogeneous of order $n$ if

$$f(t^{n}y,t^{n}p)=t^{n}f(y,p)$$

In this case, it can be seen that both $M={(-y^{2}p)}^{1/3}$ and $N=p$ are both homogeneous and of the same order $n=1$. Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution $u=\frac{p}{y}$, or $p=uy$. Hence

$$\frac{\mathrm{d}p}{\mathrm{d}y}=\frac{\mathrm{d}u}{\mathrm{d}y}y+u$$

Applying the transformation $p=uy$ to the above ODE in (1) gives

$$\begin{array}{rl}\frac{\mathrm{d}u}{\mathrm{d}y}y+u& =-\frac{1}{{(-u)}^{2/3}}\\ \frac{\mathrm{d}u}{\mathrm{d}y}& =\frac{-\frac{1}{{(-u\left(y\right))}^{2/3}}-u\left(y\right)}{y}\end{array}$$

Or

$$u^{\prime }\left(y\right)-\frac{-\frac{1}{{(-u\left(y\right))}^{2/3}}-u\left(y\right)}{y}=0$$

Or

$$u^{\prime }\left(y\right){(-u\left(y\right))}^{2/3}y+u\left(y\right){(-u\left(y\right))}^{2/3}+1=0$$

Or

$$u^{\prime }\left(y\right){(-u\left(y\right))}^{2/3}y-{(-u\left(y\right))}^{5/3}+1=0$$

Which is now solved as separable in $u\left(y\right)$.

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }\left(y\right)=\frac{{(-u\left(y\right))}^{5/3}-1}{{(-u\left(y\right))}^{2/3}y}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }\left(y\right)& =\frac{{(-u\left(y\right))}^{5/3}-1}{{(-u\left(y\right))}^{2/3}y}\\ & =f(y)g(u)\end{array}$$

Where

$$\begin{array}{rl}f(y)& =\frac{1}{y}\\ g(u)& =\frac{{(-u)}^{5/3}-1}{{(-u)}^{2/3}}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(y)dy\\ \int \frac{{(-u)}^{2/3}}{{(-u)}^{5/3}-1}du& =\int \frac{1}{y}dy\end{array}$$

$$-\frac{3\ln ({(-u\left(y\right))}^{5/3}-1)}{5}=\ln \left(y\right)+c_1$$

Taking the exponential of both sides the solution becomes

$$\frac{1}{{({(-u\left(y\right))}^{5/3}-1)}^{3/5}}=c_{1}y$$

We now need to find the singular solutions, these are found by finding for what values $g(u)$ is zero, since we had to divide by this above. Solving $g(u)=0$ or

$$\frac{{(-u)}^{5/3}-1}{{(-u)}^{2/3}}=0$$

for $u\left(y\right)$ gives

$$\begin{array}{rl}u\left(y\right)& =-1\end{array}$$

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

$$\begin{array}{rl}\frac{1}{{({(-u\left(y\right))}^{5/3}-1)}^{3/5}}& =c_{1}y\\ u\left(y\right)& =-1\end{array}$$

Converting $\frac{1}{{({(-u\left(y\right))}^{5/3}-1)}^{3/5}}=c_{1}y$ back to $p$ gives

$$\begin{array}{r}\frac{1}{{(-\frac{p{(-\frac{p}{y})}^{2/3}+y}{y})}^{3/5}}=c_{1}y\end{array}$$

Converting $u\left(y\right)=-1$ back to $p$ gives

$$\begin{array}{r}p=-y\end{array}$$

Solving Eq. (2)

Writing the ode as

$$\begin{array}{rl}{p}^{\prime }& =-\frac{{(-y^{2}p)}^{1/3}(1+i\sqrt{3})}{2p}\\ p^{\prime }& =\omega (y,p)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

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

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

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

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

$$\begin{array}{}\text{(5E)}& b_2-\frac{{(-y^{2}p)}^{1/3}(1+i\sqrt{3})(b_3-a_2)}{2p}-\frac{{(-y^{2}p)}^{2/3}{(1+i\sqrt{3})}^2{a}_3}{4p^2}-\frac{(1+i\sqrt{3})y(p{a}_3+y{a}_2+a_1)}{3{(-y^{2}p)}^{2/3}}-(\frac{(1+i\sqrt{3})y^2}{6{(-y^{2}p)}^{2/3}p}+\frac{{(-y^{2}p)}^{1/3}(1+i\sqrt{3})}{2p^2})(p{b}_3+y{b}_2+b_1)=0\end{array}$$

Putting the above in normal form gives

$$-\frac{3i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3+2i\sqrt{3}{p}^{3}y{a}_3+5i\sqrt{3}{p}^2{y}^2{a}_2-5i\sqrt{3}{p}^2{y}^2{b}_3-2i\sqrt{3}p{y}^3{b}_2+2i\sqrt{3}{p}^{2}y{a}_1-2i\sqrt{3}p{y}^2{b}_1-3{(-y^{2}p)}^{4/3}{a}_3-6b_2{(-y^{2}p)}^{2/3}{p}^2+2p^{3}y{a}_3+5p^2{y}^2{a}_2-5p^2{y}^2{b}_3-2p{y}^3{b}_2+2p^{2}y{a}_1-2p{y}^2{b}_1}{6{(-y^{2}p)}^{2/3}{p}^2}=0$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& -3i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3-2i\sqrt{3}{p}^{3}y{a}_3-5i\sqrt{3}{p}^2{y}^2{a}_2+5i\sqrt{3}{p}^2{y}^2{b}_3+2i\sqrt{3}p{y}^3{b}_2-2i\sqrt{3}{p}^{2}y{a}_1+2i\sqrt{3}p{y}^2{b}_1+3{(-y^{2}p)}^{4/3}{a}_3+6b_2{(-y^{2}p)}^{2/3}{p}^2-2p^{3}y{a}_3-5p^2{y}^2{a}_2+5p^2{y}^2{b}_3+2p{y}^3{b}_2-2p^{2}y{a}_1+2p{y}^2{b}_1=0\end{array}$$

Simplifying the above gives

$$\begin{array}{}\text{(6E)}& 6{(-y^{2}p)}^{4/3}{a}_3-6i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3-10i\sqrt{3}{p}^2{y}^2{a}_2+10i\sqrt{3}{p}^2{y}^2{b}_3+4i\sqrt{3}p{y}^3{b}_2-4i\sqrt{3}{p}^{3}y{a}_3+4i\sqrt{3}p{y}^2{b}_1-4i\sqrt{3}{p}^{2}y{a}_1-10p^2{y}^2{a}_2+10p^2{y}^2{b}_3+4p{y}^3{b}_2+12b_2{(-y^{2}p)}^{2/3}{p}^2-4p^{3}y{a}_3+4p{y}^2{b}_1-4p^{2}y{a}_1=0\end{array}$$

Since the PDE has radicals, simplifying gives

$$2p(3i{(-y^{2}p)}^{1/3}\sqrt{3}{y}^2{a}_3-2i\sqrt{3}{p}^{2}y{a}_3-5i\sqrt{3}p{y}^2{a}_2+5i\sqrt{3}p{y}^2{b}_3+2i\sqrt{3}{y}^3{b}_2-2i\sqrt{3}py{a}_1+2i\sqrt{3}{y}^2{b}_1+6{(-y^{2}p)}^{2/3}p{b}_2-3{(-y^{2}p)}^{1/3}{y}^2{a}_3-2p^{2}y{a}_3-5p{y}^2{a}_2+5p{y}^2{b}_3+2y^3{b}_2-2py{a}_1+2y^2{b}_1)=0$$

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

$$\{p,y,{(-y^{2}p)}^{1/3},{(-y^{2}p)}^{2/3}\}$$

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

$$\{p=v_1,y=v_2,{(-y^{2}p)}^{1/3}=v_3,{(-y^{2}p)}^{2/3}=v_4\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& 2v_1(3i{v}_3\sqrt{3}{v}_2^2{a}_3-2i\sqrt{3}{v}_1^2{v}_2{a}_3-5i\sqrt{3}{v}_1{v}_2^2{a}_2+5i\sqrt{3}{v}_1{v}_2^2{b}_3+2i\sqrt{3}{v}_2^3{b}_2-2i\sqrt{3}{v}_1{v}_2{a}_1+2i\sqrt{3}{v}_2^2{b}_1+6v_4{v}_1{b}_2-3v_3{v}_2^2{a}_3-2v_1^2{v}_2{a}_3-5v_1{v}_2^2{a}_2+5v_1{v}_2^2{b}_3+2v_2^3{b}_2-2v_1{v}_2{a}_1+2v_2^2{b}_1)=0\end{array}$$

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

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

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& (-4i\sqrt{3}{a}_3-4a_3)v_2{v}_1^3+(-10i\sqrt{3}{a}_2+10i\sqrt{3}{b}_3-10a_2+10b_3)v_2^2{v}_1^2+(-4i\sqrt{3}{a}_1-4a_1)v_2{v}_1^2+12b_2{v}_4{v}_1^2+(4i\sqrt{3}{b}_2+4b_2)v_2^3{v}_1+(6i\sqrt{3}{a}_3-6a_3)v_2^2{v}_3{v}_1+(4i\sqrt{3}{b}_1+4b_1)v_2^2{v}_1=0\end{array}$$

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

$$\begin{array}{rl}12b_2& =0\\ -4i\sqrt{3}{a}_1-4a_1& =0\\ -4i\sqrt{3}{a}_3-4a_3& =0\\ 4i\sqrt{3}{b}_1+4b_1& =0\\ 4i\sqrt{3}{b}_2+4b_2& =0\\ 6i\sqrt{3}{a}_3-6a_3& =0\\ -10i\sqrt{3}{a}_2+10i\sqrt{3}{b}_3-10a_2+10b_3& =0\end{array}$$

Solving the above equations for the unknowns gives

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

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

$$\begin{array}{rl}\xi & =y\\ \eta & =p\end{array}$$

Shifting is now applied to make $\xi =0$ in order to simplify the rest of the computation

$$\begin{array}{rl}\eta & =\eta -\omega (y,p)\xi \\ & =p-(-\frac{{(-y^{2}p)}^{1/3}(1+i\sqrt{3})}{2p})\left(y\right)\\ & =\frac{i{(-y^{2}p)}^{1/3}\sqrt{3}y+y{(-y^{2}p)}^{1/3}+2p^2}{2p}\\ \xi & =0\end{array}$$

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

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

The above comes from the requirements that $(\xi \frac{\partial }{\partial y}+\eta \frac{\partial }{\partial p})S(y,p)=1$. 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 $S(R)$. Since $\xi =0$ then in this special case

$$\begin{array}{r}R=y\end{array}$$

$S$ is found from

$$\begin{array}{rl}S& =\int \frac{1}{\eta }dy\\ & =\int \frac{1}{\frac{i{(-y^{2}p)}^{1/3}\sqrt{3}y+y{(-y^{2}p)}^{1/3}+2p^2}{2p}}dy\end{array}$$

Which results in

$$\begin{array}{rl}S& =\frac{3\ln (i\sqrt{3}{y}^5+2{(-y^{2}p)}^{5/3}+y^5)}{5}\end{array}$$

Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_y+\omega (y,p)S_p}{R_y+\omega (y,p)R_p}\end{array}$$

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

$$\begin{array}{rl}\omega (y,p)& =-\frac{{(-y^{2}p)}^{1/3}(1+i\sqrt{3})}{2p}\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_y& =1\\ R_p& =0\\ S_y& =\frac{-2p^{5/3}(\sqrt{3}+i)y^{1/3}+3y^2(-i+\sqrt{3})}{y(-p^{5/3}(\sqrt{3}+i)y^{1/3}+y^2(-i+\sqrt{3}))}\\ S_p& =\frac{y^{1/3}{p}^{2/3}(\sqrt{3}+i)}{p^{5/3}(\sqrt{3}+i)y^{1/3}+(i-\sqrt{3})y^2}\end{array}$$

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

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =\frac{2i{(-y^{2}p)}^{1/3}{y}^{4/3}-2p^2{y}^{1/3}(\sqrt{3}+i)+3p^{1/3}{y}^2(-i+\sqrt{3})}{p^{1/3}(-p^{5/3}(\sqrt{3}+i)y^{1/3}+y^2(-i+\sqrt{3}))y}\end{array}$$

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

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{32i-32\sqrt{3}}{R{(\sqrt{3}+i)}^5}\end{array}$$

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 $f(R)$.

$$\begin{array}{rl}\int dS& =\int \frac{32i-32\sqrt{3}}{R{(\sqrt{3}+i)}^5}dR\\ S\left(R\right)& =2\ln \left(R\right)+c_3\end{array}$$

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

$$\begin{array}{r}-\frac{3i\pi }{10}+\frac{3\ln (y^{10/3}(\sqrt{3}+i)p^{5/3}+y^5(i-\sqrt{3}))}{5}=2\ln \left(y\right)+c_3\end{array}$$

Solving Eq. (3)

Writing the ode as

$$\begin{array}{rl}{p}^{\prime }& =\frac{{(-y^{2}p)}^{1/3}(-1+i\sqrt{3})}{2p}\\ p^{\prime }& =\omega (y,p)\end{array}$$

The condition of Lie symmetry is the linearized PDE given by

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

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

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

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

$$\begin{array}{}\text{(5E)}& b_2+\frac{{(-y^{2}p)}^{1/3}(-1+i\sqrt{3})(b_3-a_2)}{2p}-\frac{{(-y^{2}p)}^{2/3}{(-1+i\sqrt{3})}^2{a}_3}{4p^2}+\frac{(-1+i\sqrt{3})y(p{a}_3+y{a}_2+a_1)}{3{(-y^{2}p)}^{2/3}}-(-\frac{(-1+i\sqrt{3})y^2}{6{(-y^{2}p)}^{2/3}p}-\frac{{(-y^{2}p)}^{1/3}(-1+i\sqrt{3})}{2p^2})(p{b}_3+y{b}_2+b_1)=0\end{array}$$

Putting the above in normal form gives

$$\frac{3i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3+2i\sqrt{3}{p}^{3}y{a}_3+5i\sqrt{3}{p}^2{y}^2{a}_2-5i\sqrt{3}{p}^2{y}^2{b}_3-2i\sqrt{3}p{y}^3{b}_2+2i\sqrt{3}{p}^{2}y{a}_1-2i\sqrt{3}p{y}^2{b}_1+3{(-y^{2}p)}^{4/3}{a}_3+6b_2{(-y^{2}p)}^{2/3}{p}^2-2p^{3}y{a}_3-5p^2{y}^2{a}_2+5p^2{y}^2{b}_3+2p{y}^3{b}_2-2p^{2}y{a}_1+2p{y}^2{b}_1}{6{(-y^{2}p)}^{2/3}{p}^2}=0$$

Setting the numerator to zero gives

$$\begin{array}{}\text{(6E)}& 3i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3+2i\sqrt{3}{p}^{3}y{a}_3+5i\sqrt{3}{p}^2{y}^2{a}_2-5i\sqrt{3}{p}^2{y}^2{b}_3-2i\sqrt{3}p{y}^3{b}_2+2i\sqrt{3}{p}^{2}y{a}_1-2i\sqrt{3}p{y}^2{b}_1+3{(-y^{2}p)}^{4/3}{a}_3+6b_2{(-y^{2}p)}^{2/3}{p}^2-2p^{3}y{a}_3-5p^2{y}^2{a}_2+5p^2{y}^2{b}_3+2p{y}^3{b}_2-2p^{2}y{a}_1+2p{y}^2{b}_1=0\end{array}$$

Simplifying the above gives

$$\begin{array}{}\text{(6E)}& 6{(-y^{2}p)}^{4/3}{a}_3+4i\sqrt{3}{p}^{3}y{a}_3+4i\sqrt{3}{p}^{2}y{a}_1-4i\sqrt{3}p{y}^2{b}_1+6i{(-y^{2}p)}^{4/3}\sqrt{3}{a}_3-4i\sqrt{3}p{y}^3{b}_2-10i\sqrt{3}{p}^2{y}^2{b}_3+10i\sqrt{3}{p}^2{y}^2{a}_2-10p^2{y}^2{a}_2+10p^2{y}^2{b}_3+4p{y}^3{b}_2+12b_2{(-y^{2}p)}^{2/3}{p}^2-4p^{3}y{a}_3+4p{y}^2{b}_1-4p^{2}y{a}_1=0\end{array}$$

Since the PDE has radicals, simplifying gives

$$2p(-3i{(-y^{2}p)}^{1/3}\sqrt{3}{y}^2{a}_3+2i\sqrt{3}{p}^{2}y{a}_3+5i\sqrt{3}p{y}^2{a}_2-5i\sqrt{3}p{y}^2{b}_3-2i\sqrt{3}{y}^3{b}_2+2i\sqrt{3}py{a}_1-2i\sqrt{3}{y}^2{b}_1+6{(-y^{2}p)}^{2/3}p{b}_2-3{(-y^{2}p)}^{1/3}{y}^2{a}_3-2p^{2}y{a}_3-5p{y}^2{a}_2+5p{y}^2{b}_3+2y^3{b}_2-2py{a}_1+2y^2{b}_1)=0$$

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

$$\{p,y,{(-y^{2}p)}^{1/3},{(-y^{2}p)}^{2/3}\}$$

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

$$\{p=v_1,y=v_2,{(-y^{2}p)}^{1/3}=v_3,{(-y^{2}p)}^{2/3}=v_4\}$$

The above PDE (6E) now becomes

$$\begin{array}{}\text{(7E)}& 2v_1(-3i{v}_3\sqrt{3}{v}_2^2{a}_3+2i\sqrt{3}{v}_1^2{v}_2{a}_3+5i\sqrt{3}{v}_1{v}_2^2{a}_2-5i\sqrt{3}{v}_1{v}_2^2{b}_3-2i\sqrt{3}{v}_2^3{b}_2+2i\sqrt{3}{v}_1{v}_2{a}_1-2i\sqrt{3}{v}_2^2{b}_1+6v_4{v}_1{b}_2-3v_3{v}_2^2{a}_3-2v_1^2{v}_2{a}_3-5v_1{v}_2^2{a}_2+5v_1{v}_2^2{b}_3+2v_2^3{b}_2-2v_1{v}_2{a}_1+2v_2^2{b}_1)=0\end{array}$$

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

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

Equation (7E) now becomes

$$\begin{array}{}\text{(8E)}& (4i\sqrt{3}{a}_3-4a_3)v_2{v}_1^3+(10i\sqrt{3}{a}_2-10i\sqrt{3}{b}_3-10a_2+10b_3)v_2^2{v}_1^2+(4i\sqrt{3}{a}_1-4a_1)v_2{v}_1^2+12b_2{v}_4{v}_1^2+(-4i\sqrt{3}{b}_2+4b_2)v_2^3{v}_1+(-6i\sqrt{3}{a}_3-6a_3)v_2^2{v}_3{v}_1+(-4i\sqrt{3}{b}_1+4b_1)v_2^2{v}_1=0\end{array}$$

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

$$\begin{array}{rl}12b_2& =0\\ -6i\sqrt{3}{a}_3-6a_3& =0\\ -4i\sqrt{3}{b}_1+4b_1& =0\\ -4i\sqrt{3}{b}_2+4b_2& =0\\ 4i\sqrt{3}{a}_1-4a_1& =0\\ 4i\sqrt{3}{a}_3-4a_3& =0\\ 10i\sqrt{3}{a}_2-10i\sqrt{3}{b}_3-10a_2+10b_3& =0\end{array}$$

Solving the above equations for the unknowns gives

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

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

$$\begin{array}{rl}\xi & =y\\ \eta & =p\end{array}$$

Shifting is now applied to make $\xi =0$ in order to simplify the rest of the computation

$$\begin{array}{rl}\eta & =\eta -\omega (y,p)\xi \\ & =p-\left(\frac{{(-y^{2}p)}^{1/3}(-1+i\sqrt{3})}{2p}\right)\left(y\right)\\ & =\frac{-i{(-y^{2}p)}^{1/3}\sqrt{3}y+y{(-y^{2}p)}^{1/3}+2p^2}{2p}\\ \xi & =0\end{array}$$

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

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

The above comes from the requirements that $(\xi \frac{\partial }{\partial y}+\eta \frac{\partial }{\partial p})S(y,p)=1$. 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 $S(R)$. Since $\xi =0$ then in this special case

$$\begin{array}{r}R=y\end{array}$$

$S$ is found from

$$\begin{array}{rl}S& =\int \frac{1}{\eta }dy\\ & =\int \frac{1}{\frac{-i{(-y^{2}p)}^{1/3}\sqrt{3}y+y{(-y^{2}p)}^{1/3}+2p^2}{2p}}dy\end{array}$$

Which results in

$$\begin{array}{rl}S& =\frac{3\ln (i\sqrt{3}{y}^5-2{(-y^{2}p)}^{5/3}-y^5)}{5}\end{array}$$

Now that $R,S$ are found, we need to setup the ode in these coordinates. This is done by evaluating

$$\begin{array}{}\text{(2)}& \frac{dS}{dR}& =\frac{S_y+\omega (y,p)S_p}{R_y+\omega (y,p)R_p}\end{array}$$

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

$$\begin{array}{rl}\omega (y,p)& =\frac{{(-y^{2}p)}^{1/3}(-1+i\sqrt{3})}{2p}\end{array}$$

Evaluating all the partial derivatives gives

$$\begin{array}{rl}{R}_y& =1\\ R_p& =0\\ S_y& =\frac{y^{1/3}(3y^{5/3}+2p^{5/3})}{(y^{2/3}{p}^{1/3}+y)(p^{4/3}{y}^{2/3}+p^{2/3}{y}^{4/3}-p^{1/3}{y}^{5/3}-py+y^2)}\\ S_p& =\frac{p^{2/3}{y}^{4/3}}{(y^{2/3}{p}^{1/3}+y)(p^{4/3}{y}^{2/3}+p^{2/3}{y}^{4/3}-p^{1/3}{y}^{5/3}-py+y^2)}\end{array}$$

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

$$\begin{array}{}\text{(2A)}& \frac{dS}{dR}& =-\frac{y^{1/3}((-1+i\sqrt{3})y{(-y^{2}p)}^{1/3}+4p^2+6p^{1/3}{y}^{5/3})}{2p^{1/3}(-p^{4/3}{y}^{2/3}+p^{1/3}{y}^{5/3}-p^{2/3}{y}^{4/3}+y(p-y))(y^{2/3}{p}^{1/3}+y)}\end{array}$$

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

$$\begin{array}{rl}\frac{dS}{dR}& =\frac{2}{R}\end{array}$$

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 $f(R)$.

$$\begin{array}{rl}\int dS& =\int \frac{2}{R}dR\\ S\left(R\right)& =2\ln \left(R\right)+c_5\end{array}$$

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

$$\begin{array}{r}-\frac{i\pi }{5}+\frac{3\ln \left(2\right)}{5}+\frac{3\ln (y^{2/3}{p}^{1/3}+y)}{5}+\frac{3\ln (p^{1/3}{y}^{11/3}-p^{2/3}{y}^{10/3}-y^{8/3}{p}^{4/3}+y^{3}p-y^4)}{5}=2\ln \left(y\right)+c_5\end{array}$$

For solution (1) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{1}{{(-\frac{y^{\prime }{(-\frac{y^{\prime }}{y})}^{2/3}+y}{y})}^{3/5}}=c_{1}y\end{array}$$

Solving for the derivative gives these ODE’s to solve

$$\begin{array}{}\text{(1)}& y^{\prime }& =-{\left(\frac{{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y\text{(2)}& y^{\prime }& =-{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y\text{(3)}& y^{\prime }& =-{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y\text{(4)}& y^{\prime }& =-{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y\text{(5)}& y^{\prime }& =-{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y\end{array}$$

Now each of the above is solved separately.

Solving Eq. (1)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y-\frac{1}{{\left(\frac{{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau =x+c_6$$

Singular solutions are found by solving

$$\begin{array}{rl}-{\left(\frac{{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\end{array}$$

Solving Eq. (2)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y-\frac{1}{{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau =x+c_7$$

Singular solutions are found by solving

$$\begin{array}{rl}-{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\end{array}$$

Solving Eq. (3)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y-\frac{1}{{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau =x+c_8$$

Singular solutions are found by solving

$$\begin{array}{rl}-{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}+\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\end{array}$$

Solving Eq. (4)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y-\frac{1}{{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau =x+c_9$$

Singular solutions are found by solving

$$\begin{array}{rl}-{\left(\frac{{(-\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5-\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\end{array}$$

Solving Eq. (5)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

$${\int }_{}^y-\frac{1}{{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau =x+\mathit{\text{\_C10}}$$

Singular solutions are found by solving

$$\begin{array}{rl}-{\left(\frac{{(\frac{\sqrt{5}}{4}-\frac{1}{4}-\frac{i\sqrt{2}\sqrt{5+\sqrt{5}}}{4})}^2{\left((c_{1}y+{\left(\frac{1}{c_{1}y}\right)}^{2/3})c_1^4{y}^4\right)}^{2/5}}{c_1^2{y}^2}\right)}^{3/2}y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y=-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\end{array}$$

For solution (2) found earlier, since $p=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=-y\end{array}$$

Integrating gives

$$\begin{array}{rl}\int -\frac{1}{y}dy& =dx\\ -\ln \left(y\right)& =x+\mathit{\text{\_C11}}\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}-y& =0\end{array}$$

for $y$. This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

$$\begin{array}{r}y=0\end{array}$$

Will add steps showing solving for IC soon.

Solving for $y$ from the above solution(s) gives (after possible removing of solutions that do not verify)

$$\begin{array}{rl}{\int }_{}^y-\frac{1}{{\left(\frac{{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau & =x+c_6\\ y& =0\\ y& =-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y& =-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\\ y& ={\mathrm{e}}^{-x-\mathit{\text{\_C11}}}\end{array}$$

Summary of solutions found

$$\begin{array}{rl}{\int }_{}^y-\frac{1}{{\left(\frac{{\left((c_1\tau +{\left(\frac{1}{c_1\tau }\right)}^{2/3})c_1^4{\tau }^4\right)}^{2/5}}{c_1^2{\tau }^2}\right)}^{3/2}\tau }d\tau & =x+c_6\\ y& =0\\ y& =-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=1)}^2}{c_1}\\ y& =-\frac{\mathrm{RootOf}{({\mathit{\text{\_Z}}}^4-{\mathit{\text{\_Z}}}^3+{\mathit{\text{\_Z}}}^2-\mathit{\text{\_Z}}+1,\mathrm{index}=4)}^2}{c_1}\\ y& ={\mathrm{e}}^{-x-\mathit{\text{\_C11}}}\end{array}$$

✓ Maple. Time used: 0.360 (sec). Leaf size: 126

ode:=y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$\begin{array}{rl}y& =0\\ y& =c_1\\ y& ={\mathrm{e}}^{\int \mathrm{RootOf}(x-{\int }_{}^{\mathit{\text{\_Z}}}-\frac{1}{{\mathit{\text{\_f}}}^2-{(-\mathit{\text{\_f}})}^{1/3}}d\mathit{\text{\_f}}+c_1)dx+c_2}\\ y& ={\mathrm{e}}^{\int \mathrm{RootOf}(x+2{\int }_{}^{\mathit{\text{\_Z}}}\frac{1}{i{(-\mathit{\text{\_f}})}^{1/3}\sqrt{3}+2{\mathit{\text{\_f}}}^2+{(-\mathit{\text{\_f}})}^{1/3}}d\mathit{\text{\_f}}+c_1)dx+c_2}\\ y& ={\mathrm{e}}^{\int \mathrm{RootOf}(x-2{\int }_{}^{\mathit{\text{\_Z}}}\frac{1}{i{(-\mathit{\text{\_f}})}^{1/3}\sqrt{3}-2{\mathit{\text{\_f}}}^2-{(-\mathit{\text{\_f}})}^{1/3}}d\mathit{\text{\_f}}+c_1)dx+c_2}\end{array}$$

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = -_b(_a)^2+(-_b(_a))^(1 
/3), _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = -1/2*I*(-_b(_a))^(1/3) 
*3^(1/2)-_b(_a)^2-1/2*(-_b(_a))^(1/3), _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE, diff(_b(_a),_a) = 1/2*I*(-_b(_a))^(1/3)* 
3^(1/2)-_b(_a)^2-1/2*(-_b(_a))^(1/3), _b(_a), explicit, HINT = [[1, 0]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[1, 0] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

✓ Mathematica. Time used: 2.742 (sec). Leaf size: 800

ode=y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}\text{Solution too large to show}\end{array}$$

✓ Sympy. Time used: 0.253 (sec). Leaf size: 3

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

$$y\left(x\right)=0$$