2.1.84 problem 83
Internal
problem
ID
[8222]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
83
Date
solved
:
Sunday, November 10, 2024 at 08:20:09 PM
CAS
classification
:
[_quadrature]
Solve
\begin{align*} y^{\prime }&=\sqrt {1-y^{2}} \end{align*}
Solved as first order autonomous ode
Time used: 0.247 (sec)
Integrating gives
\begin{align*} \int \frac {1}{\sqrt {-y^{2}+1}}d y &= dx\\ \arcsin \left (y \right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} \sqrt {-y^{2}+1}&= 0 \end{align*}
for \(y\) . 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*} y = -1\\ y = 1 \end{align*}
The following diagram is the phase line diagram. It classifies each of the above
equilibrium points as stable or not stable or semi-stable.
Solving for \(y\) gives
\begin{align*}
y &= \sin \left (x +c_1 \right ) \\
\end{align*}
Figure 2.170: Slope field plot
\(y^{\prime } = \sqrt {1-y^{2}}\)
Summary of solutions found
\begin{align*}
y &= \sin \left (x +c_1 \right ) \\
y &= -1 \\
y &= 1 \\
\end{align*}
Solved as first order Exact ode
Time used: 38.670 (sec)
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
\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}y} &= \left (\sqrt {-y^{2}+1}\right )\mathop {\mathrm {d}x}\\ \left (-\sqrt {-y^{2}+1}\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(x,y) &= -\sqrt {-y^{2}+1}\\ N(x,y) &= 1 \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
\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\sqrt {-y^{2}+1}\right )\\ &= \frac {y}{\sqrt {-y^{2}+1}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\) , 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 y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( \frac {y}{\sqrt {-y^{2}+1}}\right ) - \left (0 \right ) \right ) \\ &=\frac {y}{\sqrt {-y^{2}+1}} \end{align*}
Since \(A\) depends on \(y\) , 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 x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {1}{\sqrt {-y^{2}+1}}\left ( \left ( 0\right ) - \left (\frac {y}{\sqrt {-y^{2}+1}} \right ) \right ) \\ &=-\frac {y}{y^{2}-1} \end{align*}
Since \(B\) does not depend on \(x\) , 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}y}} \\ &= e^{\int -\frac {y}{y^{2}-1}\mathop {\mathrm {d}y} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-\frac {\ln \left (y -1\right )}{2}-\frac {\ln \left (y +1\right )}{2} } \\ &= \frac {1}{\sqrt {y -1}\, \sqrt {y +1}} \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}{\sqrt {y -1}\, \sqrt {y +1}}\left (-\sqrt {-y^{2}+1}\right ) \\ &= -\frac {\sqrt {-y^{2}+1}}{\sqrt {y -1}\, \sqrt {y +1}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{\sqrt {y -1}\, \sqrt {y +1}}\left (1\right ) \\ &= \frac {1}{\sqrt {y -1}\, \sqrt {y +1}} \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}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-\frac {\sqrt {-y^{2}+1}}{\sqrt {y -1}\, \sqrt {y +1}}\right ) + \left (\frac {1}{\sqrt {y -1}\, \sqrt {y +1}}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)
\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {N}\tag {2} \end{align*}
Integrating (1) w.r.t. \(x\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \overline {M}\mathop {\mathrm {d}x} \\
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\frac {\sqrt {-y^{2}+1}}{\sqrt {y -1}\, \sqrt {y +1}}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -\frac {\sqrt {-y^{2}+1}\, x}{\sqrt {y -1}\, \sqrt {y +1}}+ f(y) \\
\end{align*}
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\) . Taking derivative of equation (3) w.r.t \(y\) gives
\begin{align*}
\tag{4} \frac {\partial \phi }{\partial y} &= \frac {x y}{\sqrt {-y^{2}+1}\, \sqrt {y -1}\, \sqrt {y +1}}+\frac {\sqrt {-y^{2}+1}\, x}{2 \left (y -1\right )^{{3}/{2}} \sqrt {y +1}}+\frac {\sqrt {-y^{2}+1}\, x}{2 \sqrt {y -1}\, \left (y +1\right )^{{3}/{2}}}+f'(y) \\
&=0+f'(y) \\
\end{align*}
But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {1}{\sqrt {y -1}\, \sqrt {y +1}}\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {1}{\sqrt {y -1}\, \sqrt {y +1}} = 0+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[
f'(y) = \frac {1}{\sqrt {y -1}\, \sqrt {y +1}}
\]
Integrating the above w.r.t \(y\)
gives
\begin{align*}
\int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{\sqrt {y -1}\, \sqrt {y +1}}\right ) \mathop {\mathrm {d}y} \\
f(y) &= \frac {\sqrt {\left (y -1\right ) \left (y +1\right )}\, \ln \left (y +\sqrt {y^{2}-1}\right )}{\sqrt {y -1}\, \sqrt {y +1}}+ c_1 \\
\end{align*}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into
equation (3) gives \(\phi \)
\[
\phi = -\frac {\sqrt {-y^{2}+1}\, x}{\sqrt {y -1}\, \sqrt {y +1}}+\frac {\sqrt {\left (y -1\right ) \left (y +1\right )}\, \ln \left (y +\sqrt {y^{2}-1}\right )}{\sqrt {y -1}\, \sqrt {y +1}}+ c_1
\]
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 the constant \(c_1\) gives the solution as
\[
c_1 = -\frac {\sqrt {-y^{2}+1}\, x}{\sqrt {y -1}\, \sqrt {y +1}}+\frac {\sqrt {\left (y -1\right ) \left (y +1\right )}\, \ln \left (y +\sqrt {y^{2}-1}\right )}{\sqrt {y -1}\, \sqrt {y +1}}
\]
Figure 2.171: Slope field plot
\(y^{\prime } = \sqrt {1-y^{2}}\)
Summary of solutions found
\begin{align*}
-\frac {\sqrt {1-y^{2}}\, x}{\sqrt {y-1}\, \sqrt {y+1}}+\frac {\sqrt {\left (y-1\right ) \left (y+1\right )}\, \ln \left (y+\sqrt {y^{2}-1}\right )}{\sqrt {y-1}\, \sqrt {y+1}} &= c_1 \\
\end{align*}
Solved using Lie symmetry for first order ode
Time used: 0.414 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\sqrt {-y^{2}+1}\\ 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*}
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 {-y^{2}+1}\, \left (b_{3}-a_{2}\right )-\left (-y^{2}+1\right ) a_{3}+\frac {y \left (x b_{2}+y b_{3}+b_{1}\right )}{\sqrt {-y^{2}+1}} = 0
\end{equation}
Putting the above in normal form gives
\[
\frac {\sqrt {-y^{2}+1}\, y^{2} a_{3}+x y b_{2}+y^{2} a_{2}-a_{3} \sqrt {-y^{2}+1}+b_{2} \sqrt {-y^{2}+1}+y b_{1}-a_{2}+b_{3}}{\sqrt {-y^{2}+1}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} \sqrt {-y^{2}+1}\, y^{2} a_{3}+x y b_{2}+y^{2} a_{2}-a_{3} \sqrt {-y^{2}+1}+b_{2} \sqrt {-y^{2}+1}+y b_{1}-a_{2}+b_{3} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} \sqrt {-y^{2}+1}\, y^{2} a_{3}-\left (-y^{2}+1\right ) a_{2}+\left (-y^{2}+1\right ) b_{3}+x y b_{2}+y^{2} b_{3}-a_{3} \sqrt {-y^{2}+1}+b_{2} \sqrt {-y^{2}+1}+y b_{1} = 0
\end{equation}
Since the PDE has
radicals, simplifying gives
\[
\sqrt {-y^{2}+1}\, y^{2} a_{3}+x y b_{2}+y^{2} a_{2}-a_{3} \sqrt {-y^{2}+1}+b_{2} \sqrt {-y^{2}+1}+y b_{1}-a_{2}+b_{3} = 0
\]
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\left \{x, y, \sqrt {-y^{2}+1}\right \}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\left \{x = v_{1}, y = v_{2}, \sqrt {-y^{2}+1} = v_{3}\right \}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} v_{3} v_{2}^{2} a_{3}+v_{2}^{2} a_{2}+v_{1} v_{2} b_{2}-a_{3} v_{3}+v_{2} b_{1}+b_{2} v_{3}-a_{2}+b_{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} v_{1} v_{2} b_{2}+v_{3} v_{2}^{2} a_{3}+v_{2}^{2} a_{2}+v_{2} b_{1}+\left (-a_{3}+b_{2}\right ) v_{3}+b_{3}-a_{2} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} a_{2}&=0\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ -a_{3}+b_{2}&=0\\ b_{3}-a_{2}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=a_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0 \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 &= 1 \\
\eta &= 0 \\
\end{align*}
Shifting is now applied to make \(\xi =0\) in order to simplify the rest of
the computation
\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= 0 - \left (\sqrt {-y^{2}+1}\right ) \left (1\right ) \\ &= -\sqrt {-y^{2}+1}\\ \xi &= 0 \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 \(S(R)\) . Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-\sqrt {-y^{2}+1}}} dy \end{align*}
Which results in
\begin{align*} S&= -\arcsin \left (y \right ) \end{align*}
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 {-y^{2}+1} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= 0\\ S_{y} &= -\frac {1}{\sqrt {-y^{2}+1}} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -1\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} &= -1 \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 \(R,S\) .
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\) , then we only need to integrate \(f(R)\) .
\begin{align*} \int {dS} &= \int {-1\, dR}\\ S \left (R \right ) &= -R + c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} -\arcsin \left (y\right ) = -x +c_2 \end{align*}
Which gives
\begin{align*} y = -\sin \left (-x +c_2 \right ) \end{align*}
The following diagram shows solution curves of the original ode and how they transform in
the canonical coordinates space using the mapping shown.
Original ode in \(x,y\) coordinates
Canonical
coordinates
transformation
ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = \sqrt {-y^{2}+1}\)
\( \frac {d S}{d R} = -1\)
\(\!\begin {aligned} R&= x\\ S&= -\arcsin \left (y \right ) \end {aligned} \)
Figure 2.172: Slope field plot
\(y^{\prime } = \sqrt {1-y^{2}}\)
Summary of solutions found
\begin{align*}
y &= -\sin \left (-x +c_2 \right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {1-y \left (x \right )^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sqrt {1-y \left (x \right )^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {1-y \left (x \right )^{2}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\sqrt {1-y \left (x \right )^{2}}}d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arcsin \left (y \left (x \right )\right )=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\sin \left (x +\mathit {C1} \right ) \end {array} \]
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 : 0.007
(sec)
Leaf size : 8
dsolve ( diff ( y ( x ), x ) = (1-y(x)^2)^(1/2),
y(x),singsol=all)
\[
y = \sin \left (x +c_{1} \right )
\]
Mathematica DSolve solution
Solving time : 3.464
(sec)
Leaf size : 54
DSolve [{ D [ y [ x ], x ]== Sqrt [1- y [ x ]^2],{}},
y[x],x,IncludeSingularSolutions-> True ]
\begin{align*}
y(x)\to -\frac {\tan (x+c_1)}{\sqrt {\sec ^2(x+c_1)}} \\
y(x)\to \frac {\tan (x+c_1)}{\sqrt {\sec ^2(x+c_1)}} \\
y(x)\to -1 \\
y(x)\to 1 \\
\end{align*}