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

Mathematics
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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.
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.
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.

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