Question

Bernoulli differential equation by using substitution
3(1+t^2) dy/dx = 2ty(y^3 - 1)
Plz help. Will greatly appreciate the detailed answer.

Answer 1

\[\begin{align*}
3(1+t^2)\frac{dy}{dt}&=2ty(y^3-1)\\\\
3(1+t^2)\frac{dy}{dt}&=2ty^4-2ty\\\\
3(1+t^2)y^{-4}\frac{dy}{dt}&=2t-2ty^{-3}
\end{align*}\]
Set \(u=y^{-3}\), then \(\dfrac{du}{dt}=-3y^{-4}\). Then
\[\begin{align*}
-(1+t^2)\frac{du}{dt}&=2t-2tu\\\\
\frac{du}{dt}-\frac{2t}{1+t^2}u&=\frac{2t}{1+t^2}
\end{align*}\]
The equation is now linear in \(u\).