Question

Bernoulli Diff Eq, just to check?

Answer 1

\[\large \frac{dy}{dt} + ty^3 + \frac{y}{t} = 0\]
\[\large \frac{dy}{dt}= - ty^3 - \frac{y}{t}\]
So \(\large z = y^{1 - 3} = y^{-2}\)
\[\large \frac{dz}{dt} = -2y^{-3}\frac{dy}{dt}\]
\[\large \frac{dz}{dt} = -2y^{-3}(-ty^3-\frac{y}{t})\]
\[\large \frac{dz}{dt} = 2t + \frac{2}{t}y^{-2}\]
Which is linear when we make \(\large z = y^{-2}\)
\[\large \frac{dz}{dt} - \frac{2}{t}z = 2t\]
Integrating factor
\[\huge e^{-2\int\frac{1}{t}dt} = e^{-2ln(t)} = t^{-2}\]
so
\[\large (t^{-2})\frac{dz}{dt} - \frac{2}{t}z(t^{-2}) = 2t(t^{-2})\]
\[\large t^{-2}\frac{dz}{dt} - \frac{2}{t^{3}}z = \frac{2}{t}\]
Left hand side is a product rule so
\[\large (t^{-2}z)' = \frac{2}{t}\]
Integrate both sides
\[\large t^{-2}z = \int \frac{2}{t}dt\]
\[\large t^{-2}z = 2ln(t) + c\]
\[\large z = t^2(2ln(t) + c)\]
And since \(\large z = y^{-2}\) then \(\large y = \pm \sqrt{\frac{1}{z}}\)

So the final answer I get is

\[\large y = \pm \frac{1}{\sqrt{t^2(2ln(t) + c)}}\]

Answer 2

im not 100% sure but they look right to me

Answer 3

Thanks @cobra-strikes ^_^

Answer 4

haha np, just let me knw which ones aren't just in case im wrong ^_^