Solve the differential equation dy/dx = (x^2)(y^2) with the condition y = 3 when x = 0.

 

i get

 

y = -3/((x^3)-1)

 

so how'd i do?

Looks good to me (I just solved it myself). I'd integrate the -1 into the bottom to give:

 

[math]y = \frac{3}{1-x^3}[/math]

cheers, thanks cookie monster

Not a problem

 

For the record, here's the proof. Fairly straightforward, just apply seperation of variables:

 

[math]\int \frac{1}{y^2} \ dy = \int x^2 \ dx[/math]

 

So integrating gives:

 

[math]-\frac{1}{y} = \frac{x^3}{3} + c[/math]

 

Now apply initial conditions to get:

 

[math]\frac{1}{y} = \tfrac{1}{3}(1-x^3)[/math]

 

And from there, we get the answer.