26.18 problem 754

Internal problem ID [3995]
Internal file name [OUTPUT/3488_Sunday_June_05_2022_09_25_30_AM_76050534/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 26
Problem number: 754.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"

Maple gives the following as the ode type

[[_homogeneous, `class G`]]

\[ \boxed {{y^{\prime }}^{2}+b y=-a \,x^{2}} \] Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {-a \,x^{2}-b y} \tag {1} \\ y^{\prime }&=-\sqrt {-a \,x^{2}-b y} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Writing the ode as \begin {align*} y^{\prime }&=\sqrt {-a \,x^{2}-b y}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}

The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}

The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\sqrt {-a \,x^{2}-b y}\, \left (b_{3}-a_{2}\right )-\left (-a \,x^{2}-b y \right ) a_{3}+\frac {a x \left (x a_{2}+y a_{3}+a_{1}\right )}{\sqrt {-a \,x^{2}-b y}}+\frac {b \left (x b_{2}+y b_{3}+b_{1}\right )}{2 \sqrt {-a \,x^{2}-b y}} = 0 \end{equation} Putting the above in normal form gives \[ \frac {2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}+4 a \,x^{2} a_{2}-2 a \,x^{2} b_{3}+2 a x y a_{3}+2 a x a_{1}+b x b_{2}+2 b y a_{2}-b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}+b b_{1}}{2 \sqrt {-a \,x^{2}-b y}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}+4 a \,x^{2} a_{2}-2 a \,x^{2} b_{3}+2 a x y a_{3}+2 a x a_{1}+b x b_{2}+2 b y a_{2}-b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}+b b_{1} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}+2 a \,x^{2} a_{2}+2 a x y a_{3}-2 \left (-a \,x^{2}-b y \right ) a_{2}+2 \left (-a \,x^{2}-b y \right ) b_{3}+2 a x a_{1}+b x b_{2}+b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}+b b_{1} = 0 \end{equation} Since the PDE has radicals, simplifying gives \[ 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}+4 a \,x^{2} a_{2}-2 a \,x^{2} b_{3}+2 a x y a_{3}+2 a x a_{1}+b x b_{2}+2 b y a_{2}-b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}+b b_{1} = 0 \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {-a \,x^{2}-b y}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {-a \,x^{2}-b y} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 2 v_{3} a v_{1}^{2} a_{3}+4 a v_{1}^{2} a_{2}+2 a v_{1} v_{2} a_{3}-2 a v_{1}^{2} b_{3}+2 v_{3} b v_{2} a_{3}+2 a v_{1} a_{1}+2 b v_{2} a_{2}+b v_{1} b_{2}-b v_{2} b_{3}+b b_{1}+2 b_{2} v_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 2 v_{3} a v_{1}^{2} a_{3}+\left (4 a a_{2}-2 a b_{3}\right ) v_{1}^{2}+2 a v_{1} v_{2} a_{3}+\left (2 a a_{1}+b b_{2}\right ) v_{1}+2 v_{3} b v_{2} a_{3}+\left (2 b a_{2}-b b_{3}\right ) v_{2}+2 b_{2} v_{3}+b b_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} b b_{1}&=0\\ 2 b_{2}&=0\\ 2 a a_{3}&=0\\ 2 b a_{3}&=0\\ 4 a a_{2}-2 a b_{3}&=0\\ 2 b a_{2}-b b_{3}&=0\\ 2 a a_{1}+b b_{2}&=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}&=2 a_{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 &= x \\ \eta &= 2 y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {2 y}{x}\\ &= \frac {2 y}{x} \end {align*}

This is easily solved to give \begin {align*} y = c_{1} x^{2} \end {align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {y}{x^{2}} \end {align*}

And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{x} \end {align*}

Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (x \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_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end {align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by \begin {align*} \omega (x,y) &= \sqrt {-a \,x^{2}-b y} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {2 y}{x^{3}}\\ R_{y} &= \frac {1}{x^{2}}\\ S_{x} &= \frac {1}{x}\\ S_{y} &= 0 \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {x^{2}}{x \sqrt {-a \,x^{2}-b y}-2 y}\tag {2A} \end {align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= \frac {1}{\sqrt {-R b -a}-2 R} \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \frac {\ln \left (-2 R b -\sqrt {-R b -a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {4 \sqrt {-R b -a}-b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (-2 R b +\sqrt {-R b -a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {b +4 \sqrt {-R b -a}}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (4 R^{2}+R b +a \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {8 R +b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}+c_{1}\tag {4} \end {align*}

To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} \ln \left (x \right ) = \frac {\ln \left (-\frac {2 y b}{x^{2}}-\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {4 \sqrt {-\frac {y b}{x^{2}}-a}-b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (-\frac {2 y b}{x^{2}}+\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {b +4 \sqrt {-\frac {y b}{x^{2}}-a}}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (\frac {4 y^{2}}{x^{4}}+\frac {y b}{x^{2}}+a \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {\frac {8 y}{x^{2}}+b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}+c_{1} \end {align*}

Which simplifies to \begin {align*} \ln \left (x \right ) = \frac {\ln \left (-\frac {2 y b}{x^{2}}-\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {4 \sqrt {-\frac {y b}{x^{2}}-a}-b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (-\frac {2 y b}{x^{2}}+\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {b +4 \sqrt {-\frac {y b}{x^{2}}-a}}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (\frac {4 y^{2}}{x^{4}}+\frac {y b}{x^{2}}+a \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {\frac {8 y}{x^{2}}+b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \ln \left (x \right ) &= \frac {\ln \left (-\frac {2 y b}{x^{2}}-\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {4 \sqrt {-\frac {y b}{x^{2}}-a}-b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (-\frac {2 y b}{x^{2}}+\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {b +4 \sqrt {-\frac {y b}{x^{2}}-a}}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (\frac {4 y^{2}}{x^{4}}+\frac {y b}{x^{2}}+a \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {\frac {8 y}{x^{2}}+b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}+c_{1} \\ \end{align*}

Verification of solutions

\[ \ln \left (x \right ) = \frac {\ln \left (-\frac {2 y b}{x^{2}}-\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {4 \sqrt {-\frac {y b}{x^{2}}-a}-b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (-\frac {2 y b}{x^{2}}+\sqrt {-\frac {y b}{x^{2}}-a}\, b \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {b +4 \sqrt {-\frac {y b}{x^{2}}-a}}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}-\frac {\ln \left (\frac {4 y^{2}}{x^{4}}+\frac {y b}{x^{2}}+a \right )}{4}-\frac {b \,\operatorname {arctanh}\left (\frac {\frac {8 y}{x^{2}}+b}{\sqrt {b^{2}-16 a}}\right )}{2 \sqrt {b^{2}-16 a}}+c_{1} \] Verified OK.

Solving equation (2)

Writing the ode as \begin {align*} y^{\prime }&=-\sqrt {-a \,x^{2}-b y}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}

The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}

The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}-\sqrt {-a \,x^{2}-b y}\, \left (b_{3}-a_{2}\right )-\left (-a \,x^{2}-b y \right ) a_{3}-\frac {a x \left (x a_{2}+y a_{3}+a_{1}\right )}{\sqrt {-a \,x^{2}-b y}}-\frac {b \left (x b_{2}+y b_{3}+b_{1}\right )}{2 \sqrt {-a \,x^{2}-b y}} = 0 \end{equation} Putting the above in normal form gives \[ \frac {2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}-4 a \,x^{2} a_{2}+2 a \,x^{2} b_{3}-2 a x y a_{3}-2 a x a_{1}-b x b_{2}-2 b y a_{2}+b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}-b b_{1}}{2 \sqrt {-a \,x^{2}-b y}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}-4 a \,x^{2} a_{2}+2 a \,x^{2} b_{3}-2 a x y a_{3}-2 a x a_{1}-b x b_{2}-2 b y a_{2}+b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}-b b_{1} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}-2 a \,x^{2} a_{2}-2 a x y a_{3}+2 \left (-a \,x^{2}-b y \right ) a_{2}-2 \left (-a \,x^{2}-b y \right ) b_{3}-2 a x a_{1}-b x b_{2}-b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}-b b_{1} = 0 \end{equation} Since the PDE has radicals, simplifying gives \[ 2 \sqrt {-a \,x^{2}-b y}\, a \,x^{2} a_{3}+2 \sqrt {-a \,x^{2}-b y}\, b y a_{3}-4 a \,x^{2} a_{2}+2 a \,x^{2} b_{3}-2 a x y a_{3}-2 a x a_{1}-b x b_{2}-2 b y a_{2}+b y b_{3}+2 b_{2} \sqrt {-a \,x^{2}-b y}-b b_{1} = 0 \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {-a \,x^{2}-b y}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {-a \,x^{2}-b y} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 2 v_{3} a v_{1}^{2} a_{3}-4 a v_{1}^{2} a_{2}-2 a v_{1} v_{2} a_{3}+2 a v_{1}^{2} b_{3}+2 v_{3} b v_{2} a_{3}-2 a v_{1} a_{1}-2 b v_{2} a_{2}-b v_{1} b_{2}+b v_{2} b_{3}-b b_{1}+2 b_{2} v_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 2 v_{3} a v_{1}^{2} a_{3}+\left (-4 a a_{2}+2 a b_{3}\right ) v_{1}^{2}-2 a v_{1} v_{2} a_{3}+\left (-2 a a_{1}-b b_{2}\right ) v_{1}+2 v_{3} b v_{2} a_{3}+\left (-2 b a_{2}+b b_{3}\right ) v_{2}+2 b_{2} v_{3}-b b_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 2 b_{2}&=0\\ -2 a a_{3}&=0\\ 2 a a_{3}&=0\\ 2 b a_{3}&=0\\ -b b_{1}&=0\\ -4 a a_{2}+2 a b_{3}&=0\\ -2 b a_{2}+b b_{3}&=0\\ -2 a a_{1}-b b_{2}&=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}&=2 a_{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 &= x \\ \eta &= 2 y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating

Unable to determine ODE type.

26.18.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}+b y=-a \,x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\sqrt {-a \,x^{2}-b y}, y^{\prime }=-\sqrt {-a \,x^{2}-b y}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\sqrt {-a \,x^{2}-b y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\sqrt {-a \,x^{2}-b y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

✗ Solution by Maple

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

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 1.739 (sec). Leaf size: 581

DSolve[(y'[x])^2+a x^2+b y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} \text {Solve}\left [\text {RootSum}\left [\text {$\#$1}^4-\text {$\#$1}^3 b+2 \text {$\#$1}^2 a+\text {$\#$1} a b+a^2\&,\frac {2 \text {$\#$1}^3 \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )-2 \text {$\#$1}^3 \log (x)-\text {$\#$1}^2 b \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+\text {$\#$1}^2 b \log (x)+2 \text {$\#$1} a \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+a b \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )-2 \text {$\#$1} a \log (x)-a b \log (x)}{4 \text {$\#$1}^3-3 \text {$\#$1}^2 b+4 \text {$\#$1} a+a b}\&\right ]-\log \left (\sqrt {-b y(x)} \sqrt {-a x^2-b y(x)}+b y(x)\right )+\frac {1}{2} \log (y(x))+2 \log (x)&=c_1,y(x)\right ] \\ \text {Solve}\left [\text {RootSum}\left [\text {$\#$1}^4+\text {$\#$1}^3 b+2 \text {$\#$1}^2 a-\text {$\#$1} a b+a^2\&,\frac {-2 \text {$\#$1}^3 \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+2 \text {$\#$1}^3 \log (x)-\text {$\#$1}^2 b \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+\text {$\#$1}^2 b \log (x)-2 \text {$\#$1} a \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+a b \log \left (\text {$\#$1} x-\sqrt {-a x^2-b y(x)}+\sqrt {-b y(x)}\right )+2 \text {$\#$1} a \log (x)-a b \log (x)}{-4 \text {$\#$1}^3-3 \text {$\#$1}^2 b-4 \text {$\#$1} a+a b}\&\right ]-\log \left (\sqrt {-b y(x)} \sqrt {-a x^2-b y(x)}+b y(x)\right )+\frac {1}{2} \log (y(x))+2 \log (x)&=c_1,y(x)\right ] \\ \end{align*}