The solution of separable differential equation $2xy + 2x +(x^2-1)y'=0$ is $y = \frac{c+ x^{2}}{1- x^{2}} $, $c$ is a constant.
When separating variables we have to divide by $(1-x^{2})(y+1) $. Line $y=1$ satisfies the original equation, so that is the solution. But when we insert $x=1$ or $x =-1$ in the equation, we get $y=-1$. What does that mean? I understand that the points $(1,-1) $ and $(-1,-1)$ lie on $y=-1$, but what does it in general mean? Can a point be a solution of some differential equation? If we have any other separable differential equation and if $(x_{0}, y_{0})$ satisfies equation, do we consider that point as a solution (some particular or singular solution)?
Best Answer
$$(x^2-1)y'+2x(y+1)=0,$$ Let $y+1=z$ and $x^2\ne 1$, then $$z'+\frac{2x}{x^2-1}z=0 \implies \frac{dz}{z}=\frac{2x dx}{1-x^2} \implies \int\frac{dz}{z}=\int\frac{2x dx}{1-x^2} \implies \ln [z(1-x^2)]=C_1. $$ $$\implies z=\frac{C_2}{1-x^2} \implies y=\frac{C_2-1+x^2}{1-x^2}, x^2 \ne 1.$$ Clearly $y=-1$, for $C_2=0$. So the line $y=-1$ is not a singular solution is a part of the general solution.