Internal
problem
ID
[8154] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
16 Date
solved
:
Sunday, November 10, 2024 at 02:58:42 AM CAS
classification
:
[[_homogeneous, `class G`], _rational, [_Abel, `2nd type`, `class B`]]
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
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
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
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 {x}{4 y \,x^{3}-2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 y \,x^{3}-2 x^{2}\right )^{2}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {3}{2 x \left (2 x y -1\right )^{2}} \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 {4 y \,x^{3}-2 x^{2}}{x y +1}\left ( \left ( 0\right ) - \left (\frac {x}{4 y \,x^{3}-2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 y \,x^{3}-2 x^{2}\right )^{2}} \right ) \right ) \\ &=\frac {3 x}{2 x^{2} y^{2}+x y -1} \end{align*}
Since \(B\) depends on \(x\), it can not be used to obtain an integrating factor.We will now try a third
method to find an integrating factor. Let
\[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \]
\(R\) is now checked to see if it is a function of only \(t=xy\).
Therefore
\begin{align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (0\right )-\left (\frac {x}{4 y \,x^{3}-2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 y \,x^{3}-2 x^{2}\right )^{2}}\right )} {x\left (-\frac {-x y -1}{4 y \,x^{3}-2 x^{2}}\right ) - y\left (1\right )} \\ &= -\frac {3}{8 x^{3} y^{3}-10 x^{2} y^{2}+x y +1} \end{align*}
Replacing all powers of terms \(xy\) by \(t\) gives
\[ R = -\frac {3}{8 t^{3}-10 t^{2}+t +1} \]
Since \(R\) depends on \(t\) only, then it can be used to find
an integrating factor. Let the integrating factor be \(\mu \) then
\begin{align*} \mu &= e^{-\frac {3 \ln \left (t -1\right )}{5}-\frac {2 \ln \left (4 t +1\right )}{5}+\ln \left (2 t -1\right ) } \\ &= \frac {2 t -1}{\left (t -1\right )^{{3}/{5}} \left (4 t +1\right )^{{2}/{5}}} \end{align*}
Now \(t\) is replaced back with \(xy\) giving
\[ \mu =\frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}} \]
Multiplying \(M\) and \(N\) by this integrating factor gives 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 {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}\left (-\frac {-x y -1}{4 y \,x^{3}-2 x^{2}}\right ) \\ &= \frac {x y +1}{2 \left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}} x^{2}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}\left (1\right ) \\ &= \frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}} \end{align*}
A modified ODE is now obtained from the original ODE, which is exact and can solved. The
modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (\frac {x y +1}{2 \left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}} x^{2}}\right ) + \left (\frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}\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*}
\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 {x y +1}{2 \left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}} x^{2}}\mathop {\mathrm {d}x} \\
\tag{3} \phi &= \frac {\left (4 x y +1\right )^{{3}/{5}} \left (x y -1\right )^{{2}/{5}}}{2 x}+ 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 {6 \left (x y -1\right )^{{2}/{5}}}{5 \left (4 x y +1\right )^{{2}/{5}}}+\frac {\left (4 x y +1\right )^{{3}/{5}}}{5 \left (x y -1\right )^{{3}/{5}}}+f'(y) \\
&=\frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}+f'(y) \\
\end{align*}
But equation
(2) says that \(\frac {\partial \phi }{\partial y} = \frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}\). Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}} = \frac {2 x y -1}{\left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}}}+f'(y)
\end{equation}
Solving equation (5) for \( f'(y)\) gives
\[ f'(y) = 0 \]
Therefore
\[ f(y) = c_1 \]
Where \(c_1\) is constant of integration. Substituting this result for \(f(y)\) into
equation (3) gives \(\phi \)
\[
\phi = \frac {\left (4 x y +1\right )^{{3}/{5}} \left (x y -1\right )^{{2}/{5}}}{2 x}+ c_1
\]
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
\[
c_1 = \frac {\left (4 x y +1\right )^{{3}/{5}} \left (x y -1\right )^{{2}/{5}}}{2 x}
\]
The ode \(u^{\prime }\left (x \right ) = \frac {4 u \left (x \right )^{2}-3 u \left (x \right )-1}{2 x \left (2 u \left (x \right )-1\right )}\) is
separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= \frac {4 u \left (x \right )^{2}-3 u \left (x \right )-1}{2 x \left (2 u \left (x \right )-1\right )}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{2 x}\\ g(u) &= \frac {4 u^{2}-3 u -1}{2 u -1} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is
zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {4 u^{2}-3 u -1}{2 u -1}=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1\\ u \left (x \right )&=-{\frac {1}{4}} \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.
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}-\frac {\left (x y +1\right ) \left (b_{3}-a_{2}\right )}{2 x^{2} \left (2 x y -1\right )}-\frac {\left (x y +1\right )^{2} a_{3}}{4 x^{4} \left (2 x y -1\right )^{2}}-\left (-\frac {y}{2 x^{2} \left (2 x y -1\right )}+\frac {x y +1}{x^{3} \left (2 x y -1\right )}+\frac {\left (x y +1\right ) y}{x^{2} \left (2 x y -1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {1}{2 x \left (2 x y -1\right )}+\frac {x y +1}{x \left (2 x y -1\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
\frac {16 x^{6} y^{2} b_{2}-16 x^{5} y b_{2}-4 x^{4} y^{2} a_{2}-4 x^{4} y^{2} b_{3}-8 x^{3} y^{3} a_{3}-8 x^{3} y^{2} a_{1}-2 b_{2} x^{4}-8 x^{3} y a_{2}-8 x^{3} y b_{3}-11 x^{2} y^{2} a_{3}-6 x^{3} b_{1}-10 x^{2} y a_{1}+2 x^{2} a_{2}+2 x^{2} b_{3}+2 x y a_{3}+4 x a_{1}-a_{3}}{4 x^{4} \left (2 x y -1\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} 16 x^{6} y^{2} b_{2}-16 x^{5} y b_{2}-4 x^{4} y^{2} a_{2}-4 x^{4} y^{2} b_{3}-8 x^{3} y^{3} a_{3}-8 x^{3} y^{2} a_{1}-2 b_{2} x^{4}-8 x^{3} y a_{2}-8 x^{3} y b_{3}-11 x^{2} y^{2} a_{3}-6 x^{3} b_{1}-10 x^{2} y a_{1}+2 x^{2} a_{2}+2 x^{2} b_{3}+2 x y a_{3}+4 x a_{1}-a_{3} = 0
\end{equation}
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\{x, y\}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= -x \\
\eta &= y \\
\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 \\ &= y - \left (-\frac {x y +1}{2 x^{2} \left (2 x y -1\right )}\right ) \left (-x\right ) \\ &= \frac {4 x^{2} y^{2}-3 x y -1}{4 y \,x^{2}-2 x}\\ \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
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}{\frac {4 x^{2} y^{2}-3 x y -1}{4 y \,x^{2}-2 x}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {2 \ln \left (x y -1\right )}{5}+\frac {3 \ln \left (4 x y +1\right )}{5} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
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
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 \(f(R)\).
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-x y \left (x \right )-1}{4 x^{3} y \left (x \right )-2 x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-x y \left (x \right )-1}{4 x^{3} y \left (x \right )-2 x^{2}} \end {array} \]
Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous G<-homogeneous successful`