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.