Is My Solution to the Bernoulli Equation Correct?

[1MileCrash]

Homework Statement

t^2y' + 2ty - y^3 = 0

Homework Equations

The Attempt at a Solution

y'+(2/t)y = (1/t^2)y^3

Let v = y^-2; then dv/dy = -2y^-3(dy/dt) and (dy/dt)=(-1/2)y^3(dv/dy)

Making that sub we have:

(-1/2)y^3(dv/dy)+(2/t)y = (1/t^2)y^3

(dv/dy)+(-4/t)y^-2 = (-2/t^2)
(dv/dy)+(-4/t)v = (-2/t^2)


My integrating factor is t^-4, giving me

v = (2/5)t^-1 + Ct^4
y^-2 = (2/5)t^-1 + Ct^4
y^2 = ((2/5)t^-1 + Ct^4)^-1

y= (+/-) (((2/5)t^-1 + Ct^4)^-1)^(1/2)


Wolfram gives something pretty irreconcilable.

 

[Karnage1993]

Your solution is correct. Note that you can plug in both values for y back into the DE to see if the left hand side equals 0, which is does.

 

[1MileCrash]

Thanks for the response, much appreciated.

I'm having more trouble checking the validity of my solutions than finding them in the first place, turns into an algebraic mess.

For y' = ry-ky^2 (k r constant) I get:

y=((k/r)+ce^(-rt))^-1

And again wolfram gives something weird, and for

y' = ky - ny^3 I get

y= (+/-)sqrt((n/k)+ce^(-2kt)^-1)

Sorry again for the lack of latex, still doesn't work on my browser ( though admittedly if I knew the markup by memory without clicking the buttons it would work :/)