2.6.8 Problem 9 (b)
Internal
problem
ID
[18572]
Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929)
Section
:
Chapter
IV.
Methods
of
solution:
First
order
equations.
section
33.
Problems
at
page
91
Problem
number
:
9
(b)
Date
solved
:
Tuesday, January 28, 2025 at 12:00:41 PM
CAS
classification
:
[_separable]
Solve
\begin{align*} \sqrt {-u^{2}+1}\, v^{\prime }&=2 u \sqrt {1-v^{2}} \end{align*}
Solved as first order separable ode
Time used: 0.127 (sec)
The ode
\begin{equation}
v^{\prime } = \frac {2 u \sqrt {1-v^{2}}}{\sqrt {-u^{2}+1}}
\end{equation}
is separable as it can be written as
\begin{align*} v^{\prime }&= \frac {2 u \sqrt {1-v^{2}}}{\sqrt {-u^{2}+1}}\\ &= f(u) g(v) \end{align*}
Where
\begin{align*} f(u) &= \frac {2 u}{\sqrt {-u^{2}+1}}\\ g(v) &= \sqrt {-v^{2}+1} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(v)} \,dv} &= \int { f(u) \,du} \\
\int { \frac {1}{\sqrt {-v^{2}+1}}\,dv} &= \int { \frac {2 u}{\sqrt {-u^{2}+1}} \,du} \\
\end{align*}
\[
\arcsin \left (v\right )=-2 \sqrt {-u^{2}+1}+c_1
\]
We now need to find the singular solutions, these are found by finding for
what values \(g(v)\) is zero, since we had to divide by this above. Solving \(g(v)=0\) or
\[
\sqrt {-v^{2}+1}=0
\]
for \(v\) gives
\begin{align*} v&=-1\\ v&=1 \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.
Therefore the solutions found are
\begin{align*}
\arcsin \left (v\right ) &= -2 \sqrt {-u^{2}+1}+c_1 \\
v &= -1 \\
v &= 1 \\
\end{align*}
Solving for \(v\) gives
\begin{align*}
v &= -1 \\
v &= 1 \\
v &= \sin \left (-2 \sqrt {-u^{2}+1}+c_1 \right ) \\
\end{align*}
Summary of solutions found
\begin{align*}
v &= -1 \\
v &= 1 \\
v &= \sin \left (-2 \sqrt {-u^{2}+1}+c_1 \right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sqrt {-u^{2}+1}\, v^{\prime }=2 u \sqrt {1-v^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & v^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & v^{\prime }=\frac {2 u \sqrt {1-v^{2}}}{\sqrt {-u^{2}+1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {v^{\prime }}{\sqrt {1-v^{2}}}=\frac {2 u}{\sqrt {-u^{2}+1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {v^{\prime }}{\sqrt {1-v^{2}}}d u =\int \frac {2 u}{\sqrt {-u^{2}+1}}d u +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arcsin \left (v\right )=\frac {2 \left (u -1\right ) \left (u +1\right )}{\sqrt {-u^{2}+1}}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} v \\ {} & {} & v=\sin \left (\frac {\mathit {C1} \sqrt {-u^{2}+1}+2 u^{2}-2}{\sqrt {-u^{2}+1}}\right ) \end {array} \]
Maple trace
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
Maple dsolve solution
Solving time : 0.006 (sec)
Leaf size : 32
dsolve((-u^2+1)^(1/2)*diff(v(u),u) = 2*u*(1-v(u)^2)^(1/2),v(u),singsol=all)
\[
v = \sin \left (\frac {2 c_{1} \sqrt {-u^{2}+1}+2 u^{2}-2}{\sqrt {-u^{2}+1}}\right )
\]
Mathematica DSolve solution
Solving time : 0.257 (sec)
Leaf size : 44
DSolve[{Sqrt[1-u^2]*D[v[u],u]==2*u*Sqrt[1-v[u]^2],{}},v[u],u,IncludeSingularSolutions->True]
\begin{align*}
v(u)\to -\sin \left (2 \sqrt {1-u^2}-c_1\right ) \\
v(u)\to -1 \\
v(u)\to 1 \\
v(u)\to \text {Interval}[\{-1,1\}] \\
\end{align*}