1.48 problem 48

Internal problem ID [8085]
Book : Second order enumerated odes
Section : section 1
Problem number : 48
Date solved : Monday, October 21, 2024 at 04:48:08 PM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y {y^{\prime \prime }}^{4}+{y^{\prime }}^{2}&=0 \end{align*}

1.48.1 Solved as second order missing x ode

Time used: 48.028 (sec)

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{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} y p \left (y \right )^{4} \left (\frac {d}{d y}p \left (y \right )\right )^{4}+p \left (y \right )^{2} = 0 \end{align*}

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

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} p^{\prime }&=\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\ \tag{2} p^{\prime }&=\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\ \tag{3} p^{\prime }&=-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\ \tag{4} p^{\prime }&=-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}p} &= \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(y,p) &= -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= -\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}

Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=-\frac {p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}

Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)

\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}

Integrating (2) w.r.t. \(p\) gives

\begin{align*} \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\ \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\ \tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (1) says that \(\frac {\partial \phi }{\partial y} = -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes

\begin{equation} \tag{5} -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= -\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {2 p^{{3}/{2}}}{3}-\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {2 p^{{3}/{2}}}{3}-\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}} \]

Solving Eq. (2)

Writing the ode as

\begin{align*} p^{\prime }&=\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ p^{\prime }&= \omega \left ( y,p\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{y}+\omega \left ( \eta _{p}-\xi _{y}\right ) -\omega ^{2}\xi _{p}-\omega _{y}\xi -\omega _{p}\eta =0\tag {A} \end{align*}

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 &= p a_{3}+y a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+y b_{2}+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 {i \left (-p^{2} y^{3}\right )^{{1}/{4}} \left (b_{3}-a_{2}\right )}{p y}+\frac {\sqrt {-p^{2} y^{3}}\, a_{3}}{p^{2} y^{2}}-\left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p \,y^{2}}-\frac {3 i p y}{4 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) \left (p a_{3}+y a_{2}+a_{1}\right )-\left (-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) \left (p b_{3}+y b_{2}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {-i p^{4} y^{3} a_{3}+3 i p^{3} y^{4} a_{2}-6 i p^{3} y^{4} b_{3}-2 i p^{2} y^{5} b_{2}+4 b_{2} p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}-i p^{3} y^{3} a_{1}-2 i p^{2} y^{4} b_{1}+4 \left (-p^{2} y^{3}\right )^{{5}/{4}} a_{3}}{4 p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -i p^{4} y^{3} a_{3}+3 i p^{3} y^{4} a_{2}-6 i p^{3} y^{4} b_{3}-2 i p^{2} y^{5} b_{2}+4 b_{2} p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}-i p^{3} y^{3} a_{1}-2 i p^{2} y^{4} b_{1}+4 \left (-p^{2} y^{3}\right )^{{5}/{4}} a_{3} = 0 \end{equation}

Since the PDE has radicals, simplifying gives

\[ p^{2} y^{2} \left (-i p^{2} y a_{3}+3 i p \,y^{2} a_{2}-6 i p \,y^{2} b_{3}-2 i y^{3} b_{2}+4 \left (-p^{2} y^{3}\right )^{{3}/{4}} b_{2}-i p y a_{1}-2 i y^{2} b_{1}-4 \left (-p^{2} y^{3}\right )^{{1}/{4}} y a_{3}\right ) = 0 \]

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

\[ \left \{p, y, \left (-p^{2} y^{3}\right )^{{1}/{4}}, \left (-p^{2} y^{3}\right )^{{3}/{4}}\right \} \]

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

\[ \left \{p = v_{1}, y = v_{2}, \left (-p^{2} y^{3}\right )^{{1}/{4}} = v_{3}, \left (-p^{2} y^{3}\right )^{{3}/{4}} = v_{4}\right \} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} v_{1}^{2} v_{2}^{2} \left (3 i v_{1} v_{2}^{2} a_{2}-i v_{1}^{2} v_{2} a_{3}-2 i v_{2}^{3} b_{2}-6 i v_{1} v_{2}^{2} b_{3}-i v_{1} v_{2} a_{1}-2 i v_{2}^{2} b_{1}-4 v_{3} v_{2} a_{3}+4 v_{4} b_{2}\right ) = 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} -i v_{2}^{3} a_{3} v_{1}^{4}+\left (3 i a_{2}-6 i b_{3}\right ) v_{1}^{3} v_{2}^{4}-i a_{1} v_{1}^{3} v_{2}^{3}-2 i b_{2} v_{1}^{2} v_{2}^{5}-2 i b_{1} v_{1}^{2} v_{2}^{4}-4 a_{3} v_{3} v_{1}^{2} v_{2}^{3}+4 v_{4} b_{2} v_{1}^{2} v_{2}^{2} = 0 \end{equation}

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

\begin{align*} -2 i b_{1}&=0\\ -2 i b_{2}&=0\\ -i a_{1}&=0\\ -i a_{3}&=0\\ -4 a_{3}&=0\\ 4 b_{2}&=0\\ 3 i a_{2}-6 i b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=\frac {a_{2}}{2} \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 &= \frac {p}{2} \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( y,p\right ) \to \left ( R,S \right )\) 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 y}{\xi } &= \frac {d p}{\eta } = dS \tag {1} \end{align*}

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

\begin{align*} \frac {dp}{dy} &= \frac {\eta }{\xi }\\ &= \frac {\frac {p}{2}}{y}\\ &= \frac {p}{2 y} \end{align*}

This is easily solved to give

\begin{align*} p = c_1 \sqrt {y} \end{align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence

\begin{align*} R &= \frac {p}{\sqrt {y}} \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dy}{\xi } \\ &= \frac {dy}{y} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dy}{T}}\\ &= \ln \left (y \right ) \end{align*}

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{align*} \frac {dS}{dR} &= \frac { S_{y} + \omega (y,p) S_{p} }{ R_{y} + \omega (y,p) R_{p} }\tag {2} \end{align*}

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{align*} \omega (y,p) &= \frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{y} &= -\frac {p}{2 y^{{3}/{2}}}\\ R_{p} &= \frac {1}{\sqrt {y}}\\ S_{y} &= \frac {1}{y}\\ S_{p} &= 0 \end{align*}

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

\begin{align*} \frac {dS}{dR} &= \frac {2 \sqrt {y}\, p}{2 i \left (-p^{2} y^{3}\right )^{{1}/{4}}-p^{2}}\tag {2A} \end{align*}

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{align*} \frac {dS}{dR} &= \frac {2 R}{\left (-1+i\right ) \sqrt {2}\, \sqrt {R}-R^{2}} \end{align*}

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 \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {2 R}{i \sqrt {R}\, \sqrt {2}-\sqrt {R}\, \sqrt {2}-R^{2}}\, dR}\\ S \left (R \right ) &= -\frac {\left (4 i \sqrt {2}+4 \sqrt {2}\right ) \sqrt {2}\, \left (\ln \left (\frac {R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )\right )}{48}-\frac {\left (i \sqrt {2}-\sqrt {2}\right ) \sqrt {2}\, \left (\ln \left (\frac {R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )\right )}{12}-\frac {\ln \left (R^{6}+16\right )}{3}+\frac {2 i \arctan \left (\frac {R^{3}}{4}\right )}{3} + c_3 \end{align*}

\begin{align*} S \left (R \right )&= \left (-\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}\right )+\left (\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {R^{3}}{4}\right )}{3}-\frac {2 i \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}-\frac {2 i \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_3 -\frac {\ln \left (R^{6}+16\right )}{3} \end{align*}

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

\begin{align*} \ln \left (y \right ) = \left (-\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {\frac {p^{3}}{y^{{3}/{2}}}+2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {p^{3}}{y^{{3}/{2}}}-2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\left (\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {\frac {p^{3}}{y^{{3}/{2}}}-2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {p^{3}}{y^{{3}/{2}}}+2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {p^{3}}{4 y^{{3}/{2}}}\right )}{3}-\frac {2 i \arctan \left (\frac {\left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}-\frac {2 i \arctan \left (\frac {\left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_3 -\frac {\ln \left (\frac {p^{6}}{y^{3}}+16\right )}{3} \end{align*}

Solving Eq. (3)

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(y,p) \mathop {\mathrm {d}y}+ N(y,p) \mathop {\mathrm {d}p}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}p} &= \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\mathop {\mathrm {d}y}\\ \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \mathop {\mathrm {d}y} + \mathop {\mathrm {d}p} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(y,p) &= \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ N(y,p) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial p} = \frac {\partial N}{\partial y} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial p} &= \frac {\partial }{\partial p} \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right )\\ &= \frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial p} \neq \frac {\partial N}{\partial y}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial p} - \frac {\partial N}{\partial y} \right ) \\ &=1\left ( \left ( -\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) - \left (0 \right ) \right ) \\ &=\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \end{align*}

Since \(A\) depends on \(p\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial y} - \frac {\partial M}{\partial p} \right ) \\ &=\frac {p y}{\left (-p^{2} y^{3}\right )^{{1}/{4}}}\left ( \left ( 0\right ) - \left (-\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}-\frac {y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}} \right ) \right ) \\ &=\frac {1}{2 p} \end{align*}

Since \(B\) does not depend on \(y\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}p}} \\ &= e^{\int \frac {1}{2 p}\mathop {\mathrm {d}p} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{\frac {\ln \left (p \right )}{2} } \\ &= \sqrt {p} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \sqrt {p}\left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\right ) \\ &= \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \sqrt {p}\left (1\right ) \\ &= \sqrt {p} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \\ \left (\frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) + \left (\sqrt {p}\right ) \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (y,p\right )\)

\begin{align*} \frac {\partial \phi }{\partial y } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial p } &= \overline {N}\tag {2} \end{align*}

Integrating (2) w.r.t. \(p\) gives

\begin{align*} \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \overline {N}\mathop {\mathrm {d}p} \\ \int \frac {\partial \phi }{\partial p} \mathop {\mathrm {d}p} &= \int \sqrt {p}\mathop {\mathrm {d}p} \\ \tag{3} \phi &= \frac {2 p^{{3}/{2}}}{3}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(y\) and \(p\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (1) says that \(\frac {\partial \phi }{\partial y} = \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {\left (-p^{2} y^{3}\right )^{{1}/{4}}}{\sqrt {p}\, y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= \frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_4 \\ \end{align*}

Where \(c_4\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {2 p^{{3}/{2}}}{3}+\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}}+ c_4 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_4\) and \(c_2\) constants into the constant \(c_4\) gives the solution as

\[ c_4 = \frac {2 p^{{3}/{2}}}{3}+\frac {4 \left (-p^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {p}} \]

Solving Eq. (4)

Writing the ode as

\begin{align*} p^{\prime }&=-\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y}\\ p^{\prime }&= \omega \left ( y,p\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{y}+\omega \left ( \eta _{p}-\xi _{y}\right ) -\omega ^{2}\xi _{p}-\omega _{y}\xi -\omega _{p}\eta =0\tag {A} \end{align*}

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 &= p a_{3}+y a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+y b_{2}+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 {i \left (-p^{2} y^{3}\right )^{{1}/{4}} \left (b_{3}-a_{2}\right )}{p y}+\frac {\sqrt {-p^{2} y^{3}}\, a_{3}}{p^{2} y^{2}}-\left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p \,y^{2}}+\frac {3 i p y}{4 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) \left (p a_{3}+y a_{2}+a_{1}\right )-\left (\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p^{2} y}+\frac {i y^{2}}{2 \left (-p^{2} y^{3}\right )^{{3}/{4}}}\right ) \left (p b_{3}+y b_{2}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {i p^{4} y^{3} a_{3}-3 i p^{3} y^{4} a_{2}+6 i p^{3} y^{4} b_{3}+2 i p^{2} y^{5} b_{2}+4 b_{2} p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}+i p^{3} y^{3} a_{1}+2 i p^{2} y^{4} b_{1}+4 \left (-p^{2} y^{3}\right )^{{5}/{4}} a_{3}}{4 p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} i p^{4} y^{3} a_{3}-3 i p^{3} y^{4} a_{2}+6 i p^{3} y^{4} b_{3}+2 i p^{2} y^{5} b_{2}+4 b_{2} p^{2} y^{2} \left (-p^{2} y^{3}\right )^{{3}/{4}}+i p^{3} y^{3} a_{1}+2 i p^{2} y^{4} b_{1}+4 \left (-p^{2} y^{3}\right )^{{5}/{4}} a_{3} = 0 \end{equation}

Since the PDE has radicals, simplifying gives

\[ p^{2} y^{2} \left (i p^{2} y a_{3}-3 i p \,y^{2} a_{2}+6 i p \,y^{2} b_{3}+2 i y^{3} b_{2}+4 \left (-p^{2} y^{3}\right )^{{3}/{4}} b_{2}+i p y a_{1}+2 i y^{2} b_{1}-4 \left (-p^{2} y^{3}\right )^{{1}/{4}} y a_{3}\right ) = 0 \]

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

\[ \left \{p, y, \left (-p^{2} y^{3}\right )^{{1}/{4}}, \left (-p^{2} y^{3}\right )^{{3}/{4}}\right \} \]

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

\[ \left \{p = v_{1}, y = v_{2}, \left (-p^{2} y^{3}\right )^{{1}/{4}} = v_{3}, \left (-p^{2} y^{3}\right )^{{3}/{4}} = v_{4}\right \} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} v_{1}^{2} v_{2}^{2} \left (-3 i v_{1} v_{2}^{2} a_{2}+i v_{1}^{2} v_{2} a_{3}+2 i v_{2}^{3} b_{2}+6 i v_{1} v_{2}^{2} b_{3}+i v_{1} v_{2} a_{1}+2 i v_{2}^{2} b_{1}-4 v_{3} v_{2} a_{3}+4 v_{4} b_{2}\right ) = 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} i v_{2}^{3} a_{3} v_{1}^{4}+\left (-3 i a_{2}+6 i b_{3}\right ) v_{1}^{3} v_{2}^{4}+i a_{1} v_{1}^{3} v_{2}^{3}+2 i b_{2} v_{1}^{2} v_{2}^{5}+2 i b_{1} v_{1}^{2} v_{2}^{4}-4 a_{3} v_{3} v_{1}^{2} v_{2}^{3}+4 v_{4} b_{2} v_{1}^{2} v_{2}^{2} = 0 \end{equation}

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

\begin{align*} i a_{1}&=0\\ i a_{3}&=0\\ 2 i b_{1}&=0\\ 2 i b_{2}&=0\\ -4 a_{3}&=0\\ 4 b_{2}&=0\\ -3 i a_{2}+6 i b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=\frac {a_{2}}{2} \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 &= \frac {p}{2} \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( y,p\right ) \to \left ( R,S \right )\) 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 y}{\xi } &= \frac {d p}{\eta } = dS \tag {1} \end{align*}

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

\begin{align*} \frac {dp}{dy} &= \frac {\eta }{\xi }\\ &= \frac {\frac {p}{2}}{y}\\ &= \frac {p}{2 y} \end{align*}

This is easily solved to give

\begin{align*} p = c_1 \sqrt {y} \end{align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence

\begin{align*} R &= \frac {p}{\sqrt {y}} \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dy}{\xi } \\ &= \frac {dy}{y} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dy}{T}}\\ &= \ln \left (y \right ) \end{align*}

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{align*} \frac {dS}{dR} &= \frac { S_{y} + \omega (y,p) S_{p} }{ R_{y} + \omega (y,p) R_{p} }\tag {2} \end{align*}

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{align*} \omega (y,p) &= -\frac {i \left (-p^{2} y^{3}\right )^{{1}/{4}}}{p y} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{y} &= -\frac {p}{2 y^{{3}/{2}}}\\ R_{p} &= \frac {1}{\sqrt {y}}\\ S_{y} &= \frac {1}{y}\\ S_{p} &= 0 \end{align*}

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

\begin{align*} \frac {dS}{dR} &= -\frac {2 \sqrt {y}\, p}{2 i \left (-p^{2} y^{3}\right )^{{1}/{4}}+p^{2}}\tag {2A} \end{align*}

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{align*} \frac {dS}{dR} &= -\frac {2 R}{\left (-1+i\right ) \sqrt {2}\, \sqrt {R}+R^{2}} \end{align*}

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 \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {2 R}{i \sqrt {R}\, \sqrt {2}-\sqrt {R}\, \sqrt {2}+R^{2}}\, dR}\\ S \left (R \right ) &= -\frac {\left (-4 i \sqrt {2}-4 \sqrt {2}\right ) \sqrt {2}\, \left (\ln \left (\frac {R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )\right )}{48}-\frac {\left (-i \sqrt {2}+\sqrt {2}\right ) \sqrt {2}\, \left (\ln \left (\frac {R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )+2 \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )\right )}{12}-\frac {\ln \left (R^{6}+16\right )}{3}+\frac {2 i \arctan \left (\frac {R^{3}}{4}\right )}{3} + c_6 \end{align*}

\begin{align*} S \left (R \right )&= \left (\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}\right )+\left (-\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {R^{3}-2 R^{{3}/{2}} \sqrt {2}+4}{R^{3}+2 R^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {R^{3}}{4}\right )}{3}+\frac {2 i \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}+\frac {2 i \arctan \left (\frac {R^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_6 -\frac {\ln \left (R^{6}+16\right )}{3} \end{align*}

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

\begin{align*} \ln \left (y \right ) = \left (\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {\frac {p^{3}}{y^{{3}/{2}}}+2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {p^{3}}{y^{{3}/{2}}}-2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\left (-\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {\frac {p^{3}}{y^{{3}/{2}}}-2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {p^{3}}{y^{{3}/{2}}}+2 \left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {p^{3}}{4 y^{{3}/{2}}}\right )}{3}+\frac {2 i \arctan \left (\frac {\left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}+\frac {2 i \arctan \left (\frac {\left (\frac {p}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_6 -\frac {\ln \left (\frac {p^{6}}{y^{3}}+16\right )}{3} \end{align*}

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

\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}-\frac {4 \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_1 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\ \tag{2} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{3} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{4} y^{\prime }&=\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\ \tag{5} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{6} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{7} y^{\prime }&=\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\ \tag{8} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{9} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{10} y^{\prime }&=\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4} \\ \tag{11} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{12} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +c_7 \]

Solving Eq. (2)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +c_8 \]

Solving Eq. (3)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +c_9 \]

Solving Eq. (4)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C10} \]

Solving Eq. (5)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C11} \]

Solving Eq. (6)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C12} \]

Solving Eq. (7)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C13} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}{4}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_1 \right )^{{1}/{3}} c_1}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4} \end{align*}

Solving Eq. (8)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C14} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_1 \right )^{{1}/{3}} c_1}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4} \end{align*}

Solving Eq. (9)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C15} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_1 \right )^{{1}/{3}} c_1}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_1 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_1 \right )^{{1}/{3}}}{4}\right ) c_1}{4} \end{align*}

Solving Eq. (10)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C16} \]

Solving Eq. (11)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C17} \]

Solving Eq. (12)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_1 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C18} \]

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

\begin{align*} \frac {2 {y^{\prime }}^{{3}/{2}}}{3}+\frac {4 \left (-{y^{\prime }}^{2} y^{3}\right )^{{1}/{4}}}{3 \sqrt {y^{\prime }}} = c_4 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}{4} \\ \tag{2} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{3} y^{\prime }&={\left (-\frac {{\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{4} y^{\prime }&=\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}{4} \\ \tag{5} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{6} y^{\prime }&={\left (-\frac {{\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{7} y^{\prime }&=\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}{4} \\ \tag{8} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{9} y^{\prime }&={\left (-\frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{10} y^{\prime }&=\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}{4} \\ \tag{11} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \tag{12} y^{\prime }&={\left (-\frac {{\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, {\left (-16 i \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{1}/{3}}}{4}\right )}^{2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C19} \]

Solving Eq. (2)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C20} \]

Solving Eq. (3)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C21} \]

Solving Eq. (4)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C22} \]

Solving Eq. (5)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C23} \]

Solving Eq. (6)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C24} \]

Solving Eq. (7)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C25} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}{4}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_4 \right )^{{1}/{3}} c_4}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4} \end{align*}

Solving Eq. (8)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C26} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}{16}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_4 \right )^{{1}/{3}} c_4}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4} \end{align*}

Solving Eq. (9)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C27} \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (-16 \left (-y^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}{16}&= 0 \end{align*}

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{align*} y = \frac {3 \left (-6 c_4 \right )^{{1}/{3}} c_4}{8}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}-\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4}\\ y = \frac {3 \left (-\frac {\left (-6 c_4 \right )^{{1}/{3}}}{4}+\frac {i \sqrt {3}\, \left (-6 c_4 \right )^{{1}/{3}}}{4}\right ) c_4}{4} \end{align*}

Solving Eq. (10)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {4}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}}}d \tau = x +\textit {\_C28} \]

Solving Eq. (11)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2}}d \tau = x +\textit {\_C29} \]

Solving Eq. (12)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{y}\frac {16}{{\left (-16 i \left (-\tau ^{3}\right )^{{1}/{4}}+12 c_4 \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )^{2}}d \tau = x +\textit {\_C30} \]

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} \ln \left (y\right ) = \left (-\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}+2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}-2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\left (\frac {1}{6}-\frac {i}{6}\right ) \ln \left (\frac {\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}-2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}+2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {{y^{\prime }}^{3}}{4 y^{{3}/{2}}}\right )}{3}-\frac {2 i \arctan \left (\frac {\left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}-\frac {2 i \arctan \left (\frac {\left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_3 -\frac {\ln \left (\frac {{y^{\prime }}^{6}}{y^{3}}+16\right )}{3} \end{align*}

Unable to solve. Terminating.

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} \ln \left (y\right ) = \left (\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}+2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}-2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\left (-\frac {1}{6}+\frac {i}{6}\right ) \ln \left (\frac {\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}-2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}{\frac {{y^{\prime }}^{3}}{y^{{3}/{2}}}+2 \left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}+4}\right )+\frac {2 i \arctan \left (\frac {{y^{\prime }}^{3}}{4 y^{{3}/{2}}}\right )}{3}+\frac {2 i \arctan \left (\frac {\left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}-1\right )}{3}+\frac {2 i \arctan \left (\frac {\left (\frac {y^{\prime }}{\sqrt {y}}\right )^{{3}/{2}} \sqrt {2}}{2}+1\right )}{3}+c_6 -\frac {\ln \left (\frac {{y^{\prime }}^{6}}{y^{3}}+16\right )}{3} \end{align*}

Unable to solve. Terminating.

Will add steps showing solving for IC soon.

1.48.2 Maple step by step solution

1.48.3 Maple trace

Methods for second order ODEs:
 

1.48.4 Maple dsolve solution

Solving time : 3.087 (sec)
Leaf size : 2829

dsolve(y(x)*diff(diff(y(x),x),x)^4+diff(y(x),x)^2 = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= c_1 \\ y &= 0 \\ \int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} -\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} -x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (1+i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a}^{3} {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-2 \left (\textit {\_a} +\frac {\left (\textit {\_a} c_1 \right )^{{1}/{4}}}{2}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right ) \textit {\_a}^{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} -\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (1-i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (1-i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \left (1+i \sqrt {3}\right ) \textit {\_a}^{3} \left (-2 \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {\left (i-\sqrt {3}\right ) \textit {\_a}^{3} {\left (\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-2 \left (\textit {\_a} +\frac {\left (\textit {\_a} c_1 \right )^{{1}/{4}}}{2}\right ) \left (-2 \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right ) \textit {\_a}^{3}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}^{2}}{\sqrt {-\textit {\_a}^{3} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) {\left (-\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\textit {\_a} \right )-2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_f}}{2 i \left (-\textit {\_f}^{2}\right )^{{1}/{4}}+\textit {\_f}^{2}}d \textit {\_f} \right )+c_1 \right ) \sqrt {\textit {\_a}}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (2 i \textit {\_a} -\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}}{\sqrt {-i \textit {\_a} {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_2 &= 0 \\ \int _{}^{y}\frac {\textit {\_a}}{\sqrt {i \textit {\_a} {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} -x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (\sqrt {3}+i\right ) \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \textit {\_a} \left (\sqrt {3}+i\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \textit {\_a} \left (-i+\sqrt {3}\right ) \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 i \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\textit {\_a} \left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (2 i \textit {\_a} -\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-i \textit {\_a} {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {i \textit {\_a} {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-i+\sqrt {3}\right ) \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-i+\sqrt {3}\right ) \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (\sqrt {3}+i\right ) \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (-i \left (2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {-\left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \left (\sqrt {3}+i\right ) \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right ) \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i-\sqrt {3}\right ) \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (\sqrt {3}+i\right ) \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) {\left (i \left (-2 \textit {\_a} +\left (\textit {\_a} c_1 \right )^{{1}/{4}}\right ) \textit {\_a}^{2}\right )}^{{1}/{3}} \textit {\_a}}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}-2 \textit {\_a} \right ) \left (-2 i \textit {\_a}^{3}-\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \textit {\_a} \left (\sqrt {3}+i\right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ -\sqrt {2}\, \left (\int _{}^{y}\frac {\textit {\_a}}{\sqrt {\left (-2 i \textit {\_a}^{3}+\textit {\_a}^{2} \left (\textit {\_a} c_1 \right )^{{1}/{4}}\right )^{{1}/{3}} \textit {\_a} \left (-i+\sqrt {3}\right ) \left (i \left (\textit {\_a} c_1 \right )^{{1}/{4}}+2 \textit {\_a} \right )}}d \textit {\_a} \right )-x -c_2 &= 0 \\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\ln \left (\textit {\_a} \right )-2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_f}}{\textit {\_f}^{2}+2 \left (-\textit {\_f}^{2}\right )^{{1}/{4}}}d \textit {\_f} \right )+c_1 \right ) \sqrt {\textit {\_a}}}d \textit {\_a} -x -c_2 &= 0 \\ \end{align*}

1.48.5 Mathematica DSolve solution

Solving time : 6.004 (sec)
Leaf size : 1237

DSolve[{y[x]*D[y[x],{x,2}]^4+D[y[x],x]^2==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

Too large to display