Answer to Question #215066 in Differential Equations for Prince

Question

(x^2 - 1)dy - (y^2 - 1)dx

Answer 1

Solving the equation

$(x^2 - 1)dy - (y^2 - 1)dx=0$

We can rewrite as:

$(x^2-1)dy=(y^2-1)dx$

Dividing by $(x^2-1)(y^2-1)$ through:

$\frac{d y}{y^2-1}=\frac{dx}{x^2-1}$

Integrate both sides :

$\int \frac{1}{y^2-1} d y=\int \frac{1}{x^2-1} dx$

Evaluate the integrals:

$\frac{1}{2} \log (-y+1)-\frac{1}{2} \log (y+1)= \frac{1}{2} \log (-x+1)-\frac{1}{2} \log (x+1)+c_{1}$

where $c_{1}$ is an arbitrary constant.

Solving for y=y(x) we have

$y=-\frac{x+e^{2 c_{1}}(x-1)+1}{-x+e^{2 c_{1}}(x-1)-1}$

Simplifying the arbitrary constant, we have

$y=-\frac{x+c_{1}(x-1)+1}{-x+c_{1}(x-1)-1}$

which is the required solution to the given differential equation.