Internal
problem
ID
[9952] Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948 Section
:
Chapter
1,
Additional
non-linear
first
order Problem
number
:
974 Date
solved
:
Thursday, October 17, 2024 at 10:29:23 PM CAS
classification
:
[[_1st_order, _with_linear_symmetries], _Abel]
Solve
\begin{align*} y^{\prime }&=y^{3}-3 x^{2} y^{2}+3 y x^{4}-x^{6}+2 x \end{align*}
2.397.1 Solved as first order Abel ode
Time used: 0.445 (sec)
This is Abel first kind ODE, it has the form
\[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \]
Comparing the above to given ODE which is
\begin{align*}y^{\prime }&=y^{3}-3 x^{2} y^{2}+3 y x^{4}-x^{6}+2 x\tag {1} \end{align*}
\begin{align*} \int \frac {1}{u^{3}}d u &= dx\\ -\frac {1}{2 u^{2}}&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} u^{3}&= 0 \end{align*}
for \(u \left (x \right )\). 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*} u \left (x \right ) = 0\\ u \left (x \right ) = 0\\ u \left (x \right ) = 0 \end{align*}
Solving for \(u \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=\frac {1}{\sqrt {-2 c_1 -2 x}}\\ u \left (x \right )&=-\frac {1}{\sqrt {-2 c_1 -2 x}} \end{align*}
Now we transform the solution \(u \left (x \right ) = 0\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives
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}{x^{6}-3 y \,x^{4}+3 x^{2} y^{2}-y^{3}}} 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
\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 \(f(R)\).
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{3}-3 y \left (x \right )^{2} x^{2}+3 y \left (x \right ) x^{4}-x^{6}+2 x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{3}-3 y \left (x \right )^{2} x^{2}+3 y \left (x \right ) x^{4}-x^{6}+2 x \end {array} \]
2.397.4 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<-1st order, canonical coordinates successful`
\begin{align*}
y &= \frac {x^{2} \sqrt {-2 x +2 c_{1}}-1}{\sqrt {-2 x +2 c_{1}}} \\
y &= \frac {x^{2} \sqrt {-2 x +2 c_{1}}+1}{\sqrt {-2 x +2 c_{1}}} \\
\end{align*}