consider the following differential equation.....?

Question

y' = t*ln(y^2t) + t^2

is this DE seperable?

what are the ALL the constant solutions if there are more than one...?

the final answer should be a numerical answer

Answer 1

Having separated the diff. eqn. we find that it is very difficult or impossible to solve explicitly. However, the question asks for constant solutions. We see that
dy/dt = t^2(2lny + 1)
would be satisfied by
lny = -1/2 ====> y = e^(-1/2)
as this would make the bracket zero and dy/dt = 0 as needed.

However, if we left the original diff. eqn. as
dy/dt = t^2(ln(y^2) + 1)
and want the bracket to be zero then
ln(y^2) = -1 ===> y^2 = e^(-1) ===> y = +e^(-1/2) or -e^(-1/2)
so there are two constant solutions.

Answer 2

It is separable:

y' = t*ln(y^2t) + t^2

y' = 2*t^2*ln(y) + t^2

dy/dt= [t^2]*[2*ln(y) + 1]

dy/[2*ln(y) + 1] = [t^2]dt

∫dy/[2*ln(y) + 1] = ∫[t^2]dt

To solve the left hand integral, note that 1 = ln(e)

∫dy/[2*ln(y) + 1] = ∫dy/[ln(y^2) + ln(e)] = ∫dy/[ln(e*y^2)]

I cannot find a closed form for that integral.

Answer 3

dy/dt = t ln(y^2t) + t^2

Using the log properties, we can move the 2t down outside of the log.

dy/dt = (2t)(t)ln(y) + t^2

dy/dt = (2t^2)ln(y) + t^2

Factor out t^2,

dy/dt = (t^2) (2 ln(y) + 1 )

This is indeed separable.

1/[2ln(y) + 1] dy = t^2 dt

I'll leave you to solve this.