Internal
problem
ID
[9847] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
Additional
non-linear
first
order Problem
number
:
864 Date
solved
:
Friday, October 11, 2024 at 12:00:21 PM CAS
classification
:
[[_Abel, `2nd type`, `class C`], [_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
Solve
\begin{align*} y^{\prime }&=\frac {y \left ({\mathrm e}^{-\frac {x^{2}}{2}} x y+{\mathrm e}^{-\frac {x^{2}}{4}} x +2 y^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2 y \,{\mathrm e}^{-\frac {x^{2}}{4}}+2} \end{align*}
2.287.1 Solved using Lie symmetry for first order ode
Time used: 1.680 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {y \left ({\mathrm e}^{-\frac {x^{2}}{2}} x y +{\mathrm e}^{-\frac {x^{2}}{4}} x +2 y^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2 y \,{\mathrm e}^{-\frac {x^{2}}{4}}+2}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= 2 \\
\eta &= y x \\
\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 x - \left (\frac {y \left ({\mathrm e}^{-\frac {x^{2}}{2}} x y +{\mathrm e}^{-\frac {x^{2}}{4}} x +2 y^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2 y \,{\mathrm e}^{-\frac {x^{2}}{4}}+2}\right ) \left (2\right ) \\ &= \frac {-{\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}} x \,y^{2}-2 \,{\mathrm e}^{-\frac {3 x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} y^{3}-{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} x y +y^{2} {\mathrm e}^{-\frac {x^{2}}{4}} x +y x}{y \,{\mathrm e}^{-\frac {x^{2}}{4}}+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
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 {-{\mathrm e}^{-\frac {x^{2}}{2}} {\mathrm e}^{\frac {x^{2}}{4}} x \,y^{2}-2 \,{\mathrm e}^{-\frac {3 x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} y^{3}-{\mathrm e}^{-\frac {x^{2}}{4}} {\mathrm e}^{\frac {x^{2}}{4}} x y +y^{2} {\mathrm e}^{-\frac {x^{2}}{4}} x +y x}{y \,{\mathrm e}^{-\frac {x^{2}}{4}}+1}}} dy \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
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 \(f(R)\).
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) \left (\left ({\mathrm e}^{-\frac {x^{2}}{4}}\right )^{2} x y \left (x \right )+{\mathrm e}^{-\frac {x^{2}}{4}} x +2 y \left (x \right )^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2 y \left (x \right ) {\mathrm e}^{-\frac {x^{2}}{4}}+2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) \left (\left ({\mathrm e}^{-\frac {x^{2}}{4}}\right )^{2} x y \left (x \right )+{\mathrm e}^{-\frac {x^{2}}{4}} x +2 y \left (x \right )^{2} {\mathrm e}^{-\frac {3 x^{2}}{4}}\right ) {\mathrm e}^{\frac {x^{2}}{4}}}{2 y \left (x \right ) {\mathrm e}^{-\frac {x^{2}}{4}}+2} \end {array} \]
2.287.3 Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryingChinidifferentialorder: 1; looking for linear symmetriestryingexacttryingAbel<-Abel successful`