I'm trying to use fourth-order runge kutta method to solve the equation:
$$\frac{y-1}{2y}\frac{dy}{dx}=(3+\frac{y}{2})\cdot\frac{1-\frac{1}{x}}{\frac{3}{2}x+2}$$
boundary conditions:
$$y\to\infty\ as\ x\to 0$$
$$y\to0\ as\ x\to \infty$$
Here I chose x = 10 and x = 0.01 to be the boundary
however, I encountered difficulties in matching the critical point (1,1).
Maybe shooting method can help? Can someone teach me how to solve this.
Introduce another independent variable and divide up the terms so that there is no denominator left, $\frac{dx}{dt}=x(y-1)(3x+4)$, $\frac{dy}{dt}=2y(6+y)(x-1)$. The first factor ensures that the signs stay constant. One might introduce a denominator $1+x^2+y^2$ in both equation (which means $t$ gets changed) so that the right side is linearly bounded and the solutions do not grow too fast.
So in the system $$ \frac{dx}{dt}=\frac{x(y-1)(3x+4)}{1+x^2+y^2}\\~\\ \frac{dy}{dt}=\frac{2y(6+y)(x-1)}{1+x^2+y^2} $$ when moving to $(t,x,y)=(-∞,0,∞)$, $\dot x≈0$ and $\dot y≈-2$. On the other end, moving towards $(t,x,y)=(∞,∞,0)$ gives $\dot x \approx -3$, $\dot y≈0$, which looks incompatible. So there needs to be a simultaneous sign switch in both equations, for instance by multiplying both equations with $t$ or some sigmoid in $t$ like $\tanh(t)$. Now start experimenting.