Question

Solve bernoulli equation

3(1+x^2)dy/dx=2xy(y^3-1)

Answer 1

Given $3(1+x^2) \frac{dy}{dx}=2xy(y^3-1)$ .

$\implies \frac{dy}{y(y^3-1)} = \frac{2xdx}{3(1+x^2)}$

Now by integration on both sides, we get

$\int \frac{1}{y(y^3-1)} dy = \int \frac{2x}{3(1+x^2)} dx$

$\implies \dfrac{\ln\left(\left|y^3-1\right|\right)}{3}-\ln\left(\left|y\right|\right) = \dfrac{\ln\left(x^2+1\right)}{3}+ \frac{1}{3}\ln(c)$

$\implies \frac{y^3-1}{y^3 } = c (x^2+1)$

$\implies \frac{1}{y^3 } =1- c (x^2+1)$

$\implies y =\frac{1}{(1- c (x^2+1))^{\frac{1}{3}}}$ is the solution.