Answer to Question #93807 in Differential Equations for chaelven

Question

y(y²+1)dx+x(y²−1)dy=0

Answer 1

 Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:

$$x(y^2-1)dy=-y(y^2+1)dx$$ $$(y^2-1)dy=-\frac{y(y^2+1)dx}{x}$$ $$\frac{y^2-1}{y(y^2+1)}dy=-\frac{dx}{x}$$

Integrate both sides of the equation :

$$\int\frac{y^2-1}{y(y^2+1)}dy=-\int\frac{dx}{x}$$

Solve the first integral using partial fractions:

$$\frac{y^2-1}{y(y^2+1)}=\frac{A}{y}+\frac{By+C}{y^2+1}=\frac{A(y^2+1)+(By+C)y}{y(y^2+1)}$$

Then

$$y^2-1=A(y^2+1)+(By+C)y$$ $$y^2-1=Ay^2+A+By^2 +Cy$$ $$y^2-1=(A+B)y^2+A+Cy$$ $$\begin{cases} A+B=1 \\ A=-1 \\ C=0 \end{cases}$$ $$\begin{cases} A=-1\\ B=2 \\C=0 \end{cases}$$

Hence

$$\int\frac{y^2-1}{y(y^2+1)}dy=\int(-\frac{1}{y}+\frac{2y}{y^2+1})dy=$$

$$=-\int\frac{dy}{y}+\int\frac{d(y^2+1)}{y^2+1}=-ln y+ln (y^2+1)$$

Then the solution of the differential equation is:

$$-ln y+ln (y^2+1)=-\int\frac{dx}{x}$$ $$ln\frac{y^2+1}{y}=-ln{x}+ln{C}$$ $$ln\frac{y^2+1}{y}=ln\frac{C}{x}$$ $$\frac{y^2+1}{y}=\frac{C}{x}$$

Solve for y:

$$y^2+1=\frac{C}{x}y$$ $$y^2-\frac{C}{x}y+1=0$$

$$y=\frac{\frac{C}{x}\pm\sqrt{\frac{C^2}{x^2}-4}}{2}$$ $$y=\frac{\frac{C}{x}\pm\frac{\sqrt{C^2-4x^2}}{x}}{2}$$ $$y=\frac{C\pm\sqrt{C^2-4x^2}}{2x}$$

Answer: $y=\frac{C\pm\sqrt{C^2-4x^2}}{2x}$