2.5.15 Problem 15
Internal
problem
ID
[8976]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
5.0
Problem
number
:
15
Date
solved
:
Friday, February 21, 2025 at 09:01:55 PM
CAS
classification
:
[_quadrature]
Solve
\begin{align*} h^{2}+\frac {2 a h}{\sqrt {1+{h^{\prime }}^{2}}}&=b^{2} \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} h^{\prime }&=-\frac {\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}{\left (h+b \right ) \left (h-b \right )} \\
\tag{2} h^{\prime }&=\frac {\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}{\left (h+b \right ) \left (h-b \right )} \\
\end{align*}
Now each of the above is solved separately.
Solving Eq. (1)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{h}-\frac {\left (\tau +b \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_3 \]
Singular solutions are found by solving
\begin{align*} -\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (h +b \right ) \left (h -b \right )}&= 0 \end{align*}
for \(h\). 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*} h = -a -\sqrt {a^{2}+b^{2}}\\ h = -a +\sqrt {a^{2}+b^{2}}\\ h = a -\sqrt {a^{2}+b^{2}}\\ h = a +\sqrt {a^{2}+b^{2}} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((-\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (h +b \right ) \left (h -b \right )})\) is zero. These give
\begin{align*}
h&=-a -\sqrt {a^{2}+b^{2}} \\
h&=-a +\sqrt {a^{2}+b^{2}} \\
h&=a -\sqrt {a^{2}+b^{2}} \\
h&=a +\sqrt {a^{2}+b^{2}} \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
The solution \(h = -a -\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.
The solution \(h = -a +\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.
The solution \(h = a -\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.
The solution \(h = a +\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.
Solving Eq. (2)
Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as
\[ \int _{}^{h}\frac {\left (\tau +b \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_4 \]
Singular solutions are found by solving
\begin{align*} \frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (h +b \right ) \left (h -b \right )}&= 0 \end{align*}
for \(h\). 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*} h = -a -\sqrt {a^{2}+b^{2}}\\ h = -a +\sqrt {a^{2}+b^{2}}\\ h = a -\sqrt {a^{2}+b^{2}}\\ h = a +\sqrt {a^{2}+b^{2}} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((\frac {\sqrt {4 a^{2} h^{2}-b^{4}+2 h^{2} b^{2}-h^{4}}}{\left (h +b \right ) \left (h -b \right )})\) is zero. These give
\begin{align*}
h&=-a -\sqrt {a^{2}+b^{2}} \\
h&=-a +\sqrt {a^{2}+b^{2}} \\
h&=a -\sqrt {a^{2}+b^{2}} \\
h&=a +\sqrt {a^{2}+b^{2}} \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
The solution \(h = -a -\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.
The solution \(h = -a +\sqrt {a^{2}+b^{2}}\) satisfies the ode and initial conditions.
The solution \(h = a -\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.
The solution \(h = a +\sqrt {a^{2}+b^{2}}\) does not satisfiy the ode and initial conditions. Hence will not be used.
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & h \left (u \right )^{2}+\frac {2 a h \left (u \right )}{\sqrt {1+\left (\frac {d}{d u}h \left (u \right )\right )^{2}}}=b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d u}h \left (u \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d u}h \left (u \right )=\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}, \frac {d}{d u}h \left (u \right )=-\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d u}h \left (u \right )=\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}d u =\int 1d u +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=u +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d u}h \left (u \right )=-\frac {\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}{\left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {\left (\frac {d}{d u}h \left (u \right )\right ) \left (h \left (u \right )+b \right ) \left (h \left (u \right )-b \right )}{\sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}d u =\int \left (-1\right )d u +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-u +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=-u +\mathit {C1} , \frac {2 b^{4} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )-\mathit {EllipticE}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}\, \left (4 a^{2}+2 b^{2}+4 a \sqrt {a^{2}+b^{2}}\right )}-\frac {b^{2} \sqrt {1+\frac {h \left (u \right )^{2} \left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right )}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h \left (u \right )^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \left (u \right ) \sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}, \sqrt {-1+\frac {\left (4 a^{2}+2 b^{2}\right ) \left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right )}{b^{4}}}\right )}{\sqrt {-\frac {2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}}{b^{4}}}\, \sqrt {-h \left (u \right )^{4}+4 a^{2} h \left (u \right )^{2}+2 h \left (u \right )^{2} b^{2}-b^{4}}}=u +\mathit {C1} \right \} \end {array} \]
Maple trace
`Methods for first order ODEs:
-> Solving 1st order ODE of high degree, 1st attempt
trying 1st order WeierstrassP solution for high degree ODE
trying 1st order WeierstrassPPrime solution for high degree ODE
trying 1st order JacobiSN solution for high degree ODE
trying 1st order ODE linearizable_by_differentiation
trying differential order: 1; missing variables
<- differential order: 1; missing x successful`
Maple dsolve solution
Solving time : 0.485 (sec)
Leaf size : 103
dsolve(h(u)^2+2*a*h(u)/(1+diff(h(u),u)^2)^(1/2) = b^2,h(u),singsol=all)
\begin{align*}
u -\int _{}^{h \left (u \right )}\frac {\textit {\_a}^{2}-b^{2}}{\sqrt {-\textit {\_a}^{4}+\left (4 a^{2}+2 b^{2}\right ) \textit {\_a}^{2}-b^{4}}}d \textit {\_a} -c_{1} &= 0 \\
u +\int _{}^{h \left (u \right )}\frac {\textit {\_a}^{2}-b^{2}}{\sqrt {-\textit {\_a}^{4}+\left (4 a^{2}+2 b^{2}\right ) \textit {\_a}^{2}-b^{4}}}d \textit {\_a} -c_{1} &= 0 \\
\end{align*}
✓Mathematica DSolve solution
Solving time : 35.557 (sec)
Leaf size : 913
DSolve[{h[u]^2 + 2*a*h[u]/Sqrt[1 + (D[ h[u],u])^2] == b^2,{}},h[u],u,IncludeSingularSolutions->True]
\begin{align*}
h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][-u+c_1] \\
h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][u+c_1] \\
h(u)\to -\sqrt {a^2+b^2}-a \\
h(u)\to \sqrt {a^2+b^2}-a \\
\end{align*}
✗Sympy solution
Solving time : 0.000 (sec)
Leaf size : 0
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
from sympy import *
u = symbols("u")
a = symbols("a")
b = symbols("b")
h = Function("h")
ode = Eq(2*a*h(u)/sqrt(Derivative(h(u), u)**2 + 1) - b**2 + h(u)**2,0)
ics = {}
dsolve(ode,func=h(u),ics=ics)
Timed Out