3.3.9.1 Example 1

$$y^{\prime }=\frac{8y^2+12xy-10y-6x+3}{y^2+6xy-2y+9x^2-6x+1}$$

Using methods in earlier sections it can be shown that this is not isobaric for any degree including $m=1$ (which means it is not even homogeneous ode of class A, which is special case of isobaric). Let

$$\begin{array}{rl}x& =X+x_0\\ y& =Y+y_0\end{array}$$

The above ode becomes

$$\begin{array}{rl}\text{(1)}& Y^{\prime }& =\frac{8{(Y+y_0)}^2+12(X+x_0)(Y+y_0)-10(Y+y_0)-6(X+x_0)+3}{{(Y+y_0)}^2+6(X+x_0)(Y+y_0)-2(Y+y_0)+9{(X+x_0)}^2-6(X+x_0)+1}& =F(X,Y)\end{array}$$

The question now becomes how to find $x_0,y_0$ such that the above ode is isobaric of degree $1$. (i.e. homogeneous ode of class A). Earlier section showed that this becomes the condition that

$$\begin{array}{}\text{(2)}& m=-\frac{X\frac{\partial F}{\partial X}}{Y\frac{\partial F}{\partial Y}}\end{array}$$

Where $m=1$. Applying the above to (1) and setting $m=1$ gives

$$\begin{array}{rl}1& =-\frac{X\frac{d}{dX}\left(\frac{8{(Y+y_0)}^2+12(X+x_0)(Y+y_0)-10(Y+y_0)-6(X+x_0)+3}{{(Y+y_0)}^2+6(X+x_0)(Y+y_0)-2(Y+y_0)+9{(X+x_0)}^2-6(X+x_0)+1}\right)}{Y\frac{d}{dY}\left(\frac{8{(Y+y_0)}^2+12(X+x_0)(Y+y_0)-10(Y+y_0)-6(X+x_0)+3}{{(Y+y_0)}^2+6(X+x_0)(Y+y_0)-2(Y+y_0)+9{(X+x_0)}^2-6(X+x_0)+1}\right)}\\ & =-\frac{X\left(\frac{-6(3X+3Y+3x_0+3y_0-2)(2Y+2y_0-1)}{{(y_0-1+3x_0+Y+3X)}^3}\right)}{Y\left(\frac{2(3X+3Y+3x_0+3y_0-2)(6X+6x_0-1)}{{(y_0-1+3x_0+Y+3X)}^3}\right)}\\ & =-\frac{X(-6(3X+3Y+3x_0+3y_0-2)(2Y+2y_0-1))}{Y\left(2(3X+3Y+3x_0+3y_0-2)(6X+6x_0-1)\right)}\\ 1& =3\frac{X}{Y}\frac{2Y+2y_0-1}{6X+6x_0-1}\end{array}$$

The above is satisfied if $\frac{2Y+2y_0-1}{6X+6x_0-1}=\frac{1}{3}\frac{Y}{X}$. Which means $\frac{6Y+6y_0-3}{6X+6x_0-1}=\frac{Y}{X}$. This implies if $6y_0-3=0$ and $6x_0-1=0$ then the equation is satisfied. Therefore a solution is found which is

$$\begin{array}{rl}6y_0-3& =0\\ y_0& =\frac{1}{2}\end{array}$$

And

$$\begin{array}{rl}6x_0-1& =0\\ x_0& =\frac{1}{6}\end{array}$$

Since transformation is found, then substituting the above 2 equations back in (1) gives

$$\begin{array}{rl}{Y}^{\prime }& =\frac{8{(Y+\frac{1}{2})}^2+12(X+\frac{1}{6})(Y+\frac{1}{2})-10(Y+\frac{1}{2})-6(X+\frac{1}{6})+3}{{(Y+\frac{1}{2})}^2+6(X+\frac{1}{6})(Y+\frac{1}{2})-2(Y+\frac{1}{2})+9{(X+\frac{1}{6})}^2-6(X+\frac{1}{6})+1}\\ & =4\frac{3XY+2Y^2}{{(3X+Y)}^2}\\ & =G(X,Y)\end{array}$$

The above ode is now homogeneous ode of class A. We can verify this using method from above section as follows

$$\begin{array}{rl}m& =-\frac{X\frac{\partial G}{\partial X}}{Y\frac{\partial G}{\partial Y}}\\ & =\frac{-X\frac{d}{dX}\left(4\frac{Y(3X+2Y)}{{(3X+Y)}^2}\right)}{Y\frac{d}{dY}\left(4\frac{Y(3X+2Y)}{{(3X+Y)}^2}\right)}\\ & =\frac{-X(-36\frac{Y}{{(3X+Y)}^3}(X+Y))}{Y\left(36\frac{X}{{(3X+Y)}^3}(X+Y)\right)}\\ & =1\end{array}$$

We see that this is indeed homogeneous ode of class A. Now this is solved easily using the substitution $Y=uX$. This results in

$$\begin{array}{}\text{(3)}& -\ln \left(\frac{Y+X}{X}\right)+3\ln \left(\frac{Y}{X}\right)-3\ln (-\frac{3X-Y}{X})-\ln X=c_1\end{array}$$

But from earlier

$$\begin{array}{rl}X& =x-x_0\\ & =x-\frac{1}{6}\\ Y& =y-y_0\\ & =y-\frac{1}{2}\end{array}$$

Hence the solution (3) in $y\left(x\right)$ now becomes

$$\begin{array}{rl}-\ln \left(\frac{y-\frac{1}{2}+x-\frac{1}{6}}{x-\frac{1}{6}}\right)+3\ln \left(\frac{y-\frac{1}{2}}{x-\frac{1}{6}}\right)-3\ln (-\frac{3(x-\frac{1}{6})-(y-\frac{1}{2})}{x-\frac{1}{6}})-\ln (x-\frac{1}{6})& =c_2\\ -\ln \left(\frac{x+y-\frac{2}{3}}{x-\frac{1}{6}}\right)+3\ln \left(\frac{6y-3}{6x-1}\right)-3\ln \left(\frac{6y-18x}{6x-1}\right)-\ln (x-\frac{1}{6})& =c_2\\ -\ln \left(\frac{6(x+y-\frac{2}{3})}{6x-1}\right)+3\ln \left(\frac{6y-3}{6x-1}\right)-3\ln \left(6\frac{y-3x}{6x-1}\right)-\ln (x-\frac{1}{6})& =c_2\end{array}$$

The above is the solution (implicit) to the original ode. The main difficulty with this method is in solving (if possible) equation (2) when $m=1$ which is

$$1=-\frac{X\frac{\partial F}{\partial X}}{Y\frac{\partial F}{\partial Y}}$$

For $x_0,y_0$. In other words, to find explicit values for $x_0,y_0$ which makes the RHS above $1$. If we can find such $x_0,y_0$ then the original ode can now be solved. If not, then this method will not work and we say the ode is not homogeneous ode of class C. Using the software Maple this can be found as follows

restart; 
eq:=1=3*X/Y*(2*Y+2*y0-1)/(6*X+6*x0-1); 
solve(identity(eq,X),[x0,y0])
 

Which gives

[[x0 = 1/6, y0 = 1/2]]
 

And Using Mathematica

eq = 1 == 3*X/Y*(2*Y + 2*y0 - 1)/(6*X + 6*x0 - 1); 
SolveAlways[eq, {X, Y}]
 

Which gives

{{x0 -> 1/6, y0 -> 1/2}}