Internal
problem
ID
[9736] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
Additional
non-linear
first
order Problem
number
:
757 Date
solved
:
Thursday, October 17, 2024 at 09:08:28 PM CAS
classification
:
[[_1st_order, _with_linear_symmetries], _rational, [_Abel, `2nd type`, `class A`]]
Solve
\begin{align*} y^{\prime }&=\frac {-4 x y+x^{3}+2 x^{2}-4 x -8}{-8 y+2 x^{2}+4 x -8} \end{align*}
2.181.1 Solved using Lie symmetry for first order ode
Time used: 0.827 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {-x^{3}-2 x^{2}+4 x y +4 x +8}{-2 x^{2}-4 x +8 y +8}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
Substituting equations
(1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}+\frac {\left (-x^{3}-2 x^{2}+4 x y +4 x +8\right ) \left (b_{3}-a_{2}\right )}{-2 x^{2}-4 x +8 y +8}-\frac {\left (-x^{3}-2 x^{2}+4 x y +4 x +8\right )^{2} a_{3}}{4 \left (-x^{2}-2 x +4 y +4\right )^{2}}-\left (\frac {-3 x^{2}-4 x +4 y +4}{-2 x^{2}-4 x +8 y +8}-\frac {\left (-x^{3}-2 x^{2}+4 x y +4 x +8\right ) \left (-2 x -2\right )}{2 \left (-x^{2}-2 x +4 y +4\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2 x}{-x^{2}-2 x +4 y +4}-\frac {2 \left (-x^{3}-2 x^{2}+4 x y +4 x +8\right )}{\left (-x^{2}-2 x +4 y +4\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {x^{6} a_{3}+4 x^{5} a_{2}+4 x^{5} a_{3}-2 x^{5} b_{3}-6 x^{4} y a_{3}+2 x^{4} a_{1}+16 x^{4} a_{2}-4 x^{4} a_{3}-4 x^{4} b_{2}-8 x^{4} b_{3}-32 x^{3} y a_{2}-8 x^{3} y a_{3}+16 x^{3} y b_{3}+8 x^{3} a_{1}-16 x^{3} a_{2}-32 x^{3} a_{3}-16 x^{3} b_{2}+8 x^{3} b_{3}-16 x^{2} y a_{1}-64 x^{2} y a_{2}+24 x^{2} y a_{3}+32 x^{2} y b_{2}+32 x^{2} y b_{3}+64 x \,y^{2} a_{2}-32 x \,y^{2} a_{3}-32 x \,y^{2} b_{3}+32 y^{3} a_{3}-8 x^{2} a_{1}-48 x^{2} a_{2}-16 x^{2} a_{3}+16 x^{2} b_{2}+48 x^{2} b_{3}-32 x y a_{1}+128 x y a_{2}+64 x y a_{3}+64 x y b_{2}-64 x y b_{3}+32 y^{2} a_{1}+64 y^{2} a_{3}-64 y^{2} b_{2}+64 x a_{2}+64 x a_{3}+64 y a_{1}+64 y a_{2}+64 y a_{3}-128 y b_{2}-128 y b_{3}+64 a_{1}+64 a_{2}+64 a_{3}-64 b_{1}-64 b_{2}-64 b_{3}}{4 \left (x^{2}+2 x -4 y -4\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -x^{6} a_{3}-4 x^{5} a_{2}-4 x^{5} a_{3}+2 x^{5} b_{3}+6 x^{4} y a_{3}-2 x^{4} a_{1}-16 x^{4} a_{2}+4 x^{4} a_{3}+4 x^{4} b_{2}+8 x^{4} b_{3}+32 x^{3} y a_{2}+8 x^{3} y a_{3}-16 x^{3} y b_{3}-8 x^{3} a_{1}+16 x^{3} a_{2}+32 x^{3} a_{3}+16 x^{3} b_{2}-8 x^{3} b_{3}+16 x^{2} y a_{1}+64 x^{2} y a_{2}-24 x^{2} y a_{3}-32 x^{2} y b_{2}-32 x^{2} y b_{3}-64 x \,y^{2} a_{2}+32 x \,y^{2} a_{3}+32 x \,y^{2} b_{3}-32 y^{3} a_{3}+8 x^{2} a_{1}+48 x^{2} a_{2}+16 x^{2} a_{3}-16 x^{2} b_{2}-48 x^{2} b_{3}+32 x y a_{1}-128 x y a_{2}-64 x y a_{3}-64 x y b_{2}+64 x y b_{3}-32 y^{2} a_{1}-64 y^{2} a_{3}+64 y^{2} b_{2}-64 x a_{2}-64 x a_{3}-64 y a_{1}-64 y a_{2}-64 y a_{3}+128 y b_{2}+128 y b_{3}-64 a_{1}-64 a_{2}-64 a_{3}+64 b_{1}+64 b_{2}+64 b_{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
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 \\ &= x +1 - \left (\frac {-x^{3}-2 x^{2}+4 x y +4 x +8}{-2 x^{2}-4 x +8 y +8}\right ) \left (2\right ) \\ &= \frac {x^{2}+2 x -4 y +4}{x^{2}+2 x -4 y -4}\\ \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 {x^{2}+2 x -4 y +4}{x^{2}+2 x -4 y -4}}} dy \end{align*}
Which results in
\begin{align*} S&= y +2 \ln \left (-x^{2}-2 x +4 y -4\right ) \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 \(f(R)\).
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 \((R,S)\)
\( \frac {dy}{dx} = \frac {-x^{3}-2 x^{2}+4 x y +4 x +8}{-2 x^{2}-4 x +8 y +8}\)
\( \frac {d S}{d R} = \frac {R}{2}\)
\(\!\begin {aligned} R&= x\\ S&= y +2 \ln \left (-x^{2}-2 x +4 y -4\right ) \end {aligned} \)
2.181.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-4 y \left (x \right ) x +x^{3}+2 x^{2}-4 x -8}{-8 y \left (x \right )+2 x^{2}+4 x -8} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-4 y \left (x \right ) x +x^{3}+2 x^{2}-4 x -8}{-8 y \left (x \right )+2 x^{2}+4 x -8} \end {array} \]
2.181.3 Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryingChinidifferentialorder: 1; looking for linear symmetriesdifferentialorder: 1; found: 1 linear symmetries. Trying reduction of order1storder, trying the canonical coordinates of the invariance group-> Calling odsolve with the ODE`, diff(y(x), x) = (1/2)*x+1/2, y(x)` *** Sublevel 2 ***Methods for first order ODEs:--- Trying classification methods ---trying a quadrature<- quadrature successful<-1st order, canonical coordinates successful`