Bernoulli differential equation; substitution done.. plz help!

mzca

Bernoulli differential equation solution required. Will highly appreciate step-by-step detailed solution.

3(1+t^2) dy/dx = 2ty(y^3 - 1)

After, substituting
having..

du/dt - (2t / 1+t^2 )u = 2t / 1+t^2

romsek

MHF Helper
starting from

$\dfrac {du}{dt} - \left(\dfrac{2t}{1+t^2}\right)u=\dfrac{2t}{1+t^2}$

$(1+t^2)\dfrac {du}{dt} = 2t(u+1)$

$\dfrac {du}{u+1} = \dfrac {2t}{1+t^2}dt$

and you should be able to finish from here.

Prove It

MHF Helper
Bernoulli differential equation solution required. Will highly appreciate step-by-step detailed solution.

3(1+t^2) dy/dx = 2ty(y^3 - 1)

After, substituting
having..

du/dt - (2t / 1+t^2 )u = 2t / 1+t^2
It will be much easier if you recognise it is separable...

\displaystyle \begin{align*} 3 \left( 1 + t^2 \right) \, \frac{\mathrm{d}y}{\mathrm{d}t} &= 2t\,y\left( y^3 - 1 \right) \\ \frac{1}{y \left( y^3 - 1 \right) } \, \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{2t}{3 \left( 1 + t^2 \right) } \end{align*}

Now integrate both sides...