Internal
problem
ID
[4083] Book
:
Differential
equations,
Shepley
L.
Ross,
1964 Section
:
2.4,
page
55 Problem
number
:
7 Date
solved
:
Tuesday, March 04, 2025 at 05:24:27 PM CAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
Solve
\begin{align*} y^{\prime }&=\frac {-x +2 y+3}{2 x +y-1} \end{align*}
Solved as first order polynomial type ode
Time used: 0.434 (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 \(a_1=-1, b_1=2, c_1 =3, a_2=2, b_2=1, c_2=-1\). 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 \(\frac {a_1}{b_1}\neq \frac {a_2}{b_2}\). Hence this is case one where lines are not parallel. Using the transformation
\begin{align*} X &=x-x_0 \\ Y &=y-y_0 \end{align*}
Where the constants \(x_0,y_0\) are obtained by solving the following two linear algebraic equations
An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if
\[ f(t^n X, t^n Y)= t^n f(X,Y) \]
In this case, it can be seen that both \(M=-X +2 Y\) and \(N=2 X +Y\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence
An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if
\[ f(t^n X, t^n Y)= t^n f(X,Y) \]
In this case, it can be seen that both \(M=-X +2 Y\) and \(N=2 X +Y\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence
Substituting equations (1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} b_{2}+\frac {\left (-x +2 y +3\right ) \left (b_{3}-a_{2}\right )}{2 x +y -1}-\frac {\left (-x +2 y +3\right )^{2} a_{3}}{\left (2 x +y -1\right )^{2}}-\left (-\frac {1}{2 x +y -1}-\frac {2 \left (-x +2 y +3\right )}{\left (2 x +y -1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2}{2 x +y -1}-\frac {-x +2 y +3}{\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
\[
\frac {2 x^{2} a_{2}-x^{2} a_{3}-x^{2} b_{2}-2 x^{2} b_{3}+2 x y a_{2}+4 x y a_{3}+4 x y b_{2}-2 x y b_{3}-2 y^{2} a_{2}+y^{2} a_{3}+y^{2} b_{2}+2 y^{2} b_{3}-2 x a_{2}+6 x a_{3}-5 x b_{1}+x b_{2}+7 x b_{3}+5 y a_{1}-y a_{2}-7 y a_{3}-2 y b_{2}+6 y b_{3}+5 a_{1}+3 a_{2}-9 a_{3}+5 b_{1}+b_{2}-3 b_{3}}{\left (2 x +y -1\right )^{2}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} 2 x^{2} a_{2}-x^{2} a_{3}-x^{2} b_{2}-2 x^{2} b_{3}+2 x y a_{2}+4 x y a_{3}+4 x y b_{2}-2 x y b_{3}-2 y^{2} a_{2}+y^{2} a_{3}+y^{2} b_{2}+2 y^{2} b_{3}-2 x a_{2}+6 x a_{3}-5 x b_{1}+x b_{2}+7 x b_{3}+5 y a_{1}-y a_{2}-7 y a_{3}-2 y b_{2}+6 y b_{3}+5 a_{1}+3 a_{2}-9 a_{3}+5 b_{1}+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
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 -1 \\
\eta &= y +1 \\
\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 +1 - \left (\frac {-x +2 y +3}{2 x +y -1}\right ) \left (x -1\right ) \\ &= \frac {x^{2}+y^{2}-2 x +2 y +2}{2 x +y -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 {x^{2}+y^{2}-2 x +2 y +2}{2 x +y -1}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (x^{2}+y^{2}-2 x +2 y +2\right )}{2}+2 \arctan \left (\frac {2 y +2}{2 x -2}\right ) \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by
\begin{align*} \omega (x,y) &= \frac {-x +2 y +3}{2 x +y -1} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {x -2 y -3}{x^{2}+y^{2}-2 x +2 y +2}\\ S_{y} &= \frac {2 x +y -1}{x^{2}+y^{2}-2 x +2 y +2} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
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} &= 0 \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 \(f(R)\).
Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).
Eliminating \(p\) from the following two equations