5.15 problem 15
Internal
problem
ID
[7956]
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:
Own
collection
of
miscellaneous
problems
Section
:
section
5.0
Problem
number
:
15
Date
solved
:
Monday, October 21, 2024 at 04:39:27 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)
Since initial conditions \(\left (u_0,h_0\right ) \) 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
\[ \int _{}^{h}-\frac {\left (\tau +b \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_1 \]
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)
Since initial conditions \(\left (u_0,h_0\right ) \) 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
\[ \int _{}^{h}\frac {\left (\tau +b \right ) \left (\tau -b \right )}{\sqrt {4 a^{2} \tau ^{2}-b^{4}+2 b^{2} \tau ^{2}-\tau ^{4}}}d \tau = u +c_2 \]
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.
5.15.1 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & h^{2}+\frac {2 a h}{\sqrt {1+{h^{\prime }}^{2}}}=b^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & h^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [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 )}, 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 )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} 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 )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {h^{\prime } \left (h+b \right ) \left (h-b \right )}{\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {h^{\prime } \left (h+b \right ) \left (h-b \right )}{\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \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 \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^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \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^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=u +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} 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 )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {h^{\prime } \left (h+b \right ) \left (h-b \right )}{\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {h^{\prime } \left (h+b \right ) \left (h-b \right )}{\sqrt {-h^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \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 \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^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \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^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=-u +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {2 b^{4} \sqrt {1+\frac {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \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 \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^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \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^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=-u +\mathit {C1} , \frac {2 b^{4} \sqrt {1+\frac {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \left (\mathit {EllipticF}\left (h \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 \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^{4}+4 a^{2} h^{2}+2 h^{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 {\left (2 a \sqrt {a^{2}+b^{2}}-2 a^{2}-b^{2}\right ) h^{2}}{b^{4}}}\, \sqrt {1-\frac {\left (2 a \sqrt {a^{2}+b^{2}}+2 a^{2}+b^{2}\right ) h^{2}}{b^{4}}}\, \mathit {EllipticF}\left (h \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^{4}+4 a^{2} h^{2}+2 h^{2} b^{2}-b^{4}}}=u +\mathit {C1} \right \} \end {array} \]
5.15.2 Maple trace
Methods for first order ODEs:
5.15.3 Maple dsolve solution
Solving time : 0.348 (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 -\left (\int _{}^{h}\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} \right )-c_1 &= 0 \\
u +\int _{}^{h}\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*}
5.15.4 Mathematica DSolve solution
Solving time : 24.227 (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*}