M mzca Nov 2014 1 0 MBD Nov 24, 2014 #1 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

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 Nov 2013 6,837 3,079 California Nov 24, 2014 #2 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.

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 Aug 2008 12,897 5,001 Nov 25, 2014 #3 mzca said: 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 Click to expand... 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...

mzca said: 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 Click to expand... 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...