Question

how can you solve this equation using Bernoulli euqation:
t^2(dy/dt)+2ty-y^3=0, t>0

Answer 1

Try the method of integrating factors.
Rewrite this as \[t ^{2}y' + 2ty = y ^{3}\].
Then notice the Left Hand Side is
the derivative of \[t ^{2}y\] with respect to t.
This is because y is a function of t.
So using the product rule,
the derivative of \[t ^{2}y\] with respect to t is
\[t ^{2}y' + 2ty = y ^{3}\].

If the Left Hand Side is something, then
so is the Right Hand Side.

So the Right Hand Side is also
the derivative of \[t ^{2}y\] with respect to t.

Then integrate both sides with respect to t.

The Left Hand Side becomes just \[t ^{2}y\].
The Right Hand Side becomes
\[ty ^{3} + C\].

Then the solution to the differential equation
is the set of all curves y(t) in the y-t plane such that
\[t ^{2}y - ty ^{3} = C\] where we can vary C.

Answer 2

Sorry that was a bit sloppy, I was testing the Equation Tool.
Also there's a typo. The derivative of t^2 *y should be t^2 * y' + 2ty, without the " = y^3" part.

Answer 3

Actually, the solution is incorrect. Sorry about that.
Omnideus asked the same questions a few hours after you and caught my mistake.
The problem is that when we integrated y^3 with respect to t, we don’t get t*y^3 since y itself is a function of t.

So the method of solving Bernoulli equations actually first starts of dividing the Left and Right Hand Sides by y^3 in order to avoid that problem.

Then it goes on to using a substitution to re-write the equation so that we can use the method of integrating factors more easily.

See http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx or omnideus’s post for more details. I’m sorry about the confusion.