Internal
problem
ID
[4082] Book
:
Differential
equations,
Shepley
L.
Ross,
1964 Section
:
2.4,
page
55 Problem
number
:
6 Date
solved
:
Monday, January 27, 2025 at 08:07:50 AM CAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
Solve
\begin{align*} 3 x -y+1-\left (6 x -2 y-3\right ) y^{\prime }&=0 \end{align*}
Solved as first order polynomial type ode
Time used: 0.240 (sec)
This is ODE of type polynomial. Where the RHS of the ode is ratio of equations of two
lines. Writing the ODE in the form
\[ y^{\prime }= \frac {a_1 x + b_1 y + c_1}{ a_2 x + b_2 y + c_3 } \]
Where \(a_1=3, b_1=-1, c_1 =1, a_2=6, b_2=-2, c_2=-3\). There are now two possible solution
methods. The first case is when the two lines \(a_1 x + b_1 y + c_1\),\( a_2 x + b_2 y + c_3\) are not parallel and the second case is
if they are parallel. If they are not parallel, then the transformation \(X=x-x_0\), \(Y=y-y_0\) converts
the ODE to a homogeneous ODE. The values \( x_0,y_0\) have to be determined. If they
are parallel then a transformation \(U(x)=a_1 x + b_1 y\) converts the given ODE in \(y\) to a separable
ODE in \(U(x)\). The first case is when \(\frac {a_1}{b_1} \neq \frac {a_2}{b_2}\) and the second case when \(\frac {a_1}{b_1} = \frac {a_2}{b_2}\). From the above we
see that \(\frac {a_1}{b_1}=\frac {3}{-1}=-3\) and \(\frac {a_2}{b_2}=\frac {6}{-2}=-3\). Hence this is case two, where the lines are parallel. Let \(U(x)=3 x -y\). Solving
for \(y\) gives
\[ y=3 x -U \left (x \right ) \]
Taking derivative w.r.t \(x\) gives
\[ y^{\prime }=3-U^{\prime }\left (x \right ) \]
Substituting the above into the ODE
results in the ODE
\[ U^{\prime }\left (x \right )=\frac {5 U \left (x \right )-10}{2 U \left (x \right )-3} \]
Which is now solved as separable in \(U \left (x \right )\). Integrating gives
\begin{align*} \int \frac {2 U -3}{5 U -10}d U &= dx\\ \frac {2 U}{5}+\frac {\ln \left (U -2\right )}{5}&= x +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} \frac {5 U -10}{2 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 ) = 2 \end{align*}
Solving for \(U \left (x \right )\) gives
\begin{align*}
U \left (x \right ) &= 2 \\
U \left (x \right ) &= \frac {\operatorname {LambertW}\left (2 \,{\mathrm e}^{-4+5 x +5 c_2}\right )}{2}+2 \\
\end{align*}
Each of the above solutions is now converted converted back to \(y\). The
solution \(U \left (x \right ) = 2\) is converted to \(y\) using \(U \left (x \right ) = 3 x -y\). Which gives
\[ 3 x -y = 2 \]
Or
\[ y = -2+3 x \]
The solution \(U \left (x \right ) = \frac {\operatorname {LambertW}\left (2 \,{\mathrm e}^{-4+5 x +5 c_2}\right )}{2}+2\) is converted to \(y\) using \(U \left (x \right ) = 3 x -y\).
Which gives
\[ 3 x -y = \frac {\operatorname {LambertW}\left (2 \,{\mathrm e}^{-4+5 x +5 c_2}\right )}{2}+2 \]
Or
\[ y = -\frac {\operatorname {LambertW}\left (2 \,{\mathrm e}^{-4+5 x +5 c_2}\right )}{2}-2+3 x \]
Figure 2.11: Slope field plot \(3 x -y+1-\left (6 x -2 y-3\right ) y^{\prime } = 0\)
Summary of solutions found
\begin{align*}
y &= -2+3 x \\
y &= -\frac {\operatorname {LambertW}\left (2 \,{\mathrm e}^{-4+5 x +5 c_2}\right )}{2}-2+3 x \\
\end{align*}
Solved using Lie symmetry for first order ode
Time used: 0.547 (sec)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {y -3 x -1}{2 y -6 x +3}\\ 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 (y -3 x -1\right ) \left (b_{3}-a_{2}\right )}{2 y -6 x +3}-\frac {\left (y -3 x -1\right )^{2} a_{3}}{\left (2 y -6 x +3\right )^{2}}-\left (-\frac {3}{2 y -6 x +3}+\frac {6 y -18 x -6}{\left (2 y -6 x +3\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {1}{2 y -6 x +3}-\frac {2 \left (y -3 x -1\right )}{\left (2 y -6 x +3\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {18 x^{2} a_{2}+9 x^{2} a_{3}-36 x^{2} b_{2}-18 x^{2} b_{3}-12 x y a_{2}-6 x y a_{3}+24 x y b_{2}+12 x y b_{3}+2 y^{2} a_{2}+y^{2} a_{3}-4 y^{2} b_{2}-2 y^{2} b_{3}-18 x a_{2}+6 x a_{3}+41 x b_{2}+3 x b_{3}+y a_{2}-17 y a_{3}-12 y b_{2}+4 y b_{3}-15 a_{1}-3 a_{2}+a_{3}+5 b_{1}-9 b_{2}+3 b_{3}}{\left (-2 y +6 x -3\right )^{2}} = 0
\]
Setting the
numerator to zero gives
\begin{equation}
\tag{6E} -18 x^{2} a_{2}-9 x^{2} a_{3}+36 x^{2} b_{2}+18 x^{2} b_{3}+12 x y a_{2}+6 x y a_{3}-24 x y b_{2}-12 x y b_{3}-2 y^{2} a_{2}-y^{2} a_{3}+4 y^{2} b_{2}+2 y^{2} b_{3}+18 x a_{2}-6 x a_{3}-41 x b_{2}-3 x b_{3}-y a_{2}+17 y a_{3}+12 y b_{2}-4 y b_{3}+15 a_{1}+3 a_{2}-a_{3}-5 b_{1}+9 b_{2}-3 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 \\ &= 3 - \left (\frac {y -3 x -1}{2 y -6 x +3}\right ) \left (1\right ) \\ &= \frac {-5 y +15 x -10}{-2 y +6 x -3}\\ \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 {-5 y +15 x -10}{-2 y +6 x -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
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 \(f(R)\).
for \(p \left (x \right )\). 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*} p \left (x \right ) = 3\\ p \left (x \right ) = {\frac {1}{2}} \end{align*}
Substituing the above solution for \(p\) in (2A) gives
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x -y \left (x \right )+1-\left (6 x -2 y \left (x \right )-3\right ) \left (\frac {d}{d x}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {-3 x +y \left (x \right )-1}{6 x -2 y \left (x \right )-3} \end {array} \]
Maple trace
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous C1storder, trying the canonical coordinates of the invariance group-> Calling odsolve with the ODE`, diff(y(x), x) = 3, y(x)` *** Sublevel 2 ***Methods for first order ODEs:--- Trying classification methods ---trying a quadraturetrying 1st order linear<- 1st order linear successful<-1st order, canonical coordinates successful<-homogeneous successful`