Internal problem ID [10044]
Internal file name [OUTPUT/8991_Monday_June_06_2022_06_09_08_AM_65905338/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1723 (book 6.132).
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime } y-{y^{\prime }}^{2}-2 a y \left ({y^{\prime }}^{2}+1\right )^{\frac {3}{2}}=1} \]
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(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y -p \left (y \right )^{2}-2 a y \left (p \left (y \right )^{2}+1\right )^{\frac {3}{2}} = 1 \end {align*}
Which is now solved as first order ode for \(p(y)\).
Entering Exact first order ODE solver. (Form one type)
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*} \left (p y\right )\mathop {\mathrm {d}p} &= \left (p^{2}+2 a y \left (p^{2}+1\right )^{\frac {3}{2}}+1\right )\mathop {\mathrm {d}y}\\ \left (-p^{2}-2 a y \left (p^{2}+1\right )^{\frac {3}{2}}-1\right )\mathop {\mathrm {d}y} + \left (p y\right )\mathop {\mathrm {d}p} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(y,p) &= -p^{2}-2 a y \left (p^{2}+1\right )^{\frac {3}{2}}-1\\ N(y,p) &= p y \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 (-p^{2}-2 a y \left (p^{2}+1\right )^{\frac {3}{2}}-1\right )\\ &= -6 p \left (a y \sqrt {p^{2}+1}+\frac {1}{3}\right ) \end {align*}
And \begin {align*} \frac {\partial N}{\partial y} &= \frac {\partial }{\partial y} \left (p y\right )\\ &= p \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 ) \\ &=\frac {1}{p y}\left ( \left ( -2 p -6 a y \sqrt {p^{2}+1}\, p\right ) - \left (p \right ) \right ) \\ &=\frac {-6 a y \sqrt {p^{2}+1}-3}{y} \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 {1}{\left (p^{2}+1\right ) \left (2 a y \sqrt {p^{2}+1}+1\right )}\left ( \left ( p\right ) - \left (-2 p -6 a y \sqrt {p^{2}+1}\, p \right ) \right ) \\ &=-\frac {3 p}{p^{2}+1} \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 {3 p}{p^{2}+1}\mathop {\mathrm {d}p} } \end {align*}
The result of integrating gives \begin {align*} \mu &= e^{-\frac {3 \ln \left (p^{2}+1\right )}{2} } \\ &= \frac {1}{\left (p^{2}+1\right )^{\frac {3}{2}}} \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 \\ &= \frac {1}{\left (p^{2}+1\right )^{\frac {3}{2}}}\left (-p^{2}-2 a y \left (p^{2}+1\right )^{\frac {3}{2}}-1\right ) \\ &= \frac {-2 a y \sqrt {p^{2}+1}-1}{\sqrt {p^{2}+1}} \end {align*}
And \begin {align*} \overline {N} &=\mu N \\ &= \frac {1}{\left (p^{2}+1\right )^{\frac {3}{2}}}\left (p y\right ) \\ &= \frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}} \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 {-2 a y \sqrt {p^{2}+1}-1}{\sqrt {p^{2}+1}}\right ) + \left (\frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}}\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 (1) w.r.t. \(y\) gives \begin{align*} \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int \overline {M}\mathop {\mathrm {d}y} \\ \int \frac {\partial \phi }{\partial y} \mathop {\mathrm {d}y} &= \int \frac {-2 a y \sqrt {p^{2}+1}-1}{\sqrt {p^{2}+1}}\mathop {\mathrm {d}y} \\ \tag{3} \phi &= \frac {\left (-a y \sqrt {p^{2}+1}-1\right ) y}{\sqrt {p^{2}+1}}+ f(p) \\ \end{align*} Where \(f(p)\) 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 \(p\) gives \begin{align*} \tag{4} \frac {\partial \phi }{\partial p} &= -\frac {a \,y^{2} p}{p^{2}+1}-\frac {\left (-a y \sqrt {p^{2}+1}-1\right ) y p}{\left (p^{2}+1\right )^{\frac {3}{2}}}+f'(p) \\ &=\frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}}+f'(p) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial p} = \frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}} = \frac {p y}{\left (p^{2}+1\right )^{\frac {3}{2}}}+f'(p) \end{equation} Solving equation (5) for \( f'(p)\) gives \[ f'(p) = 0 \] Therefore \[ f(p) = c_{1} \] Where \(c_{1}\) is constant of integration. Substituting this result for \(f(p)\) into equation (3) gives \(\phi \) \[ \phi = \frac {\left (-a y \sqrt {p^{2}+1}-1\right ) y}{\sqrt {p^{2}+1}}+ c_{1} \] 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 new constant \(c_{1}\) gives the solution as \[ c_{1} = \frac {\left (-a y \sqrt {p^{2}+1}-1\right ) y}{\sqrt {p^{2}+1}} \] 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 {\left (-a y \sqrt {{y^{\prime }}^{2}+1}-1\right ) y}{\sqrt {{y^{\prime }}^{2}+1}} = c_{1} \end {align*}
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 }&=\frac {\sqrt {-a^{2} y^{4}-2 y^{2} a c_{1} +y^{2}-c_{1}^{2}}}{a y^{2}+c_{1}} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-a^{2} y^{4}-2 y^{2} a c_{1} +y^{2}-c_{1}^{2}}}{a y^{2}+c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {a \,y^{2}+c_{1}}{\sqrt {-a^{2} y^{4}-2 a c_{1} y^{2}-c_{1}^{2}+y^{2}}}d y &= \int {dx}\\ \int _{}^{y}\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a}&= c_{3} +x \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {a \,y^{2}+c_{1}}{\sqrt {-a^{2} y^{4}-2 a c_{1} y^{2}-c_{1}^{2}+y^{2}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a}&= x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} &= c_{3} +x \\ \tag{2} \int _{}^{y}-\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} = c_{3} +x \] Verified OK.
\[ \int _{}^{y}-\frac {\textit {\_a}^{2} a +c_{1}}{\sqrt {-\textit {\_a}^{4} a^{2}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} = x +c_{4} \] Verified OK.
Maple trace
`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 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(2*a*_a*(_b(_a)^2+1)^(3/2)+_b(_a)^2+1)/_a = 0, _b(_a)` *** Sublevel 2 Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 5`[0, (_b^2+1)^(3/2)/_b/_a]
✓ Solution by Maple
Time used: 0.312 (sec). Leaf size: 117
dsolve(diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2-1-2*a*y(x)*(diff(y(x),x)^2+1)^(3/2)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -i x +c_{1} \\ y \left (x \right ) &= i x +c_{1} \\ \int _{}^{y \left (x \right )}\frac {a \,\textit {\_a}^{2}+c_{1}}{\sqrt {-a^{2} \textit {\_a}^{4}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {a \,\textit {\_a}^{2}+c_{1}}{\sqrt {-a^{2} \textit {\_a}^{4}-2 \textit {\_a}^{2} a c_{1} +\textit {\_a}^{2}-c_{1}^{2}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 5.652 (sec). Leaf size: 2181
DSolve[-1 - y'[x]^2 - 2*a*y[x]*(1 + y'[x]^2)^(3/2) + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a c_1+\sqrt {1-4 a c_1}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \left (\left (-2 a c_1+\sqrt {1-4 a c_1}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right )|-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )-\left (1+\sqrt {1-4 a c_1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right ),-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a c_1)+c_1{}^2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a c_1+\sqrt {1-4 a c_1}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \left (\left (-2 a c_1+\sqrt {1-4 a c_1}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right )|-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )-\left (1+\sqrt {1-4 a c_1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right ),-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a c_1)+c_1{}^2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a (-c_1)+\sqrt {1-4 a (-c_1)}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a (-c_1)+\sqrt {1-4 a (-c_1)}-1}} \left (\left (-2 a (-c_1)+\sqrt {1-4 a (-c_1)}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}} \text {$\#$1}\right )|-\frac {2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}{-2 (-c_1) a+\sqrt {1-4 a (-c_1)}+1}\right )-\left (1+\sqrt {1-4 a (-c_1)}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}} \text {$\#$1}\right ),-\frac {2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}{-2 (-c_1) a+\sqrt {1-4 a (-c_1)}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a (-c_1)+\sqrt {1-4 a (-c_1)}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a (-c_1))+(-c_1){}^2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a (-c_1)+\sqrt {1-4 a (-c_1)}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a (-c_1)+\sqrt {1-4 a (-c_1)}-1}} \left (\left (-2 a (-c_1)+\sqrt {1-4 a (-c_1)}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}} \text {$\#$1}\right )|-\frac {2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}{-2 (-c_1) a+\sqrt {1-4 a (-c_1)}+1}\right )-\left (1+\sqrt {1-4 a (-c_1)}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}} \text {$\#$1}\right ),-\frac {2 (-c_1) a+\sqrt {1-4 a (-c_1)}-1}{-2 (-c_1) a+\sqrt {1-4 a (-c_1)}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a (-c_1)+\sqrt {1-4 a (-c_1)}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a (-c_1))+(-c_1){}^2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a c_1+\sqrt {1-4 a c_1}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \left (\left (-2 a c_1+\sqrt {1-4 a c_1}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right )|-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )-\left (1+\sqrt {1-4 a c_1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right ),-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a c_1)+c_1{}^2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {1-\frac {2 \text {$\#$1}^2 a^2}{-2 a c_1+\sqrt {1-4 a c_1}+1}} \sqrt {1+\frac {2 \text {$\#$1}^2 a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \left (\left (-2 a c_1+\sqrt {1-4 a c_1}+1\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right )|-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )-\left (1+\sqrt {1-4 a c_1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \text {$\#$1}\right ),-\frac {2 a c_1+\sqrt {1-4 a c_1}-1}{-2 a c_1+\sqrt {1-4 a c_1}+1}\right )\right )}{2 \sqrt {2} a \sqrt {\frac {a^2}{2 a c_1+\sqrt {1-4 a c_1}-1}} \sqrt {\text {$\#$1}^4 a^2+\text {$\#$1}^2 (-1+2 a c_1)+c_1{}^2}}\&\right ][x+c_2] \\ \end{align*}