Initial Value Problem

1. Sep 5, 2007

robierob12

Im having a little issue figuring out this intial value problem.

Solve the Initial Value Problem
y' = (3x^2)/[(3y^2)-4] where y(0)=1

Looks like I can solve it as a seperable DE.

dy/dx = (3x^2)/[(3y^2)-4]

[(3y^2)-4] dy = (3x^2) dx

Integrating both sides.....

(y^3) - 4y = (x^3) + c

I don't see how to get this in terms of y = (explicitly)
to find my c...

Am I just missing some easy algebra or did I use the worng method for this one?

Thanks,
Rob

2. Sep 5, 2007

Kummer

$$y' = \frac{3x^2}{3y^2-4}$$
So,
$$(3y^2-4)y' = 3x^2$$
Integrate,
$$\int (3y^2-4)y' dx = \int 3x^2 dx$$.
That means,
$$y^3 - 4y = x^3 + C$$
Now at $$x=0$$ it means $$y=1$$.
Use this to find $$C$$ and complete the problem.

3. Sep 5, 2007

robierob12

It's funny sometimes how easy something can end up being... just pluging in a point.

Thanks,
Rob