- Thread starter
- #1

#### alane1994

##### Active member

- Oct 16, 2012

- 126

*Solve the initial value problem*

\(y\prime=\dfrac{3x^2}{3y^2-4},~y(1)=0\)

*and determine the interval in which the solution if valid.*

Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.

Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.

My work so far:

\(y\prime=\dfrac{3x^2}{3y^2-4}\)

\(\dfrac{dy}{dx}=\dfrac{3x^2}{3y^2-4}\)

\((3y^2-4)dx=(3x^2)dx\)

\(\int (3y^2-4)dy=\int (3x^2)dx\)

\(y^3-4y=x^3+C,~C=-1\)

\(y^3-4y=x^3-1\)

And this is where I am at so far. If you could tell me if this is correct as of yet, and guide me to the final destination of this problem, that would be joyous!