# IVP and Interval of Definition

#### alane1994

##### Active member
Here is my question:
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.

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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Here is my question:
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.

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!
Hi alane1994!

Yes. That is correct.
You do have a typo. It should be $$(3y^2-4)d\color{red}{\mathbf y}=(3x^2)dx$$.

To find a vertical tangent you need a point where $dx/dy=0$.
I suggest you substitute $dx=0$ in $$(3y^2-4)dy=(3x^2)dx$$ and solve it.

#### alane1994

##### Active member
If $$\dfrac{dy}{dx}=0$$, does that not mean that the slope is zero and as such the line is horizontal?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If $$\dfrac{dy}{dx}=0$$, does that not mean that the slope is zero and as such the line is horizontal?
Yes.
That is why I mentioned $$\dfrac{dx}{dy}=0$$ instead (x and y swapped).

#### alane1994

##### Active member
Aha, I feel silly now, I feel very silly indeed.

#### alane1994

##### Active member
When I do that and solve for y, I get:
$$y=\pm2,~0$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
When I do that and solve for y, I get:
$$y=\pm2,~0$$
That doesn't sound right.
I don't think that is the solution of $3y^2 - 4 = 0$.

#### alane1994

##### Active member
Isn't it $$(3y^2-4)dy=0$$?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Isn't it $$(3y^2-4)dy=0$$?
Yes. And we'll assume (for now) that $dy \ne 0$.

When we substitute one of your solutions, say $y=2$, we get:
$$(3y^2-4)dy = (3 \cdot 2^2 - 4) dy = 8dy \ne 0$$
So $y=2$ is not a solution.

#### alane1994

##### Active member
But what I am saying, is that the dy means that it is a derivative of something yes? so to solve for y truly, you need to take the integral of it, and then solve for y. Right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
But what I am saying, is that the dy means that it is a derivative of something yes? so to solve for y truly, you need to take the integral of it, and then solve for y. Right?
Not really.

You can rewrite your original differential equation $$\displaystyle \frac{dy}{dx}=\frac{3x^2}{3y^2−4}$$ as $$\displaystyle \frac{dx}{dy}=\frac{3y^2−4}{3x^2}$$ due to the Inverse function theorem. After that you can equate it to 0 and solve it, yielding vertical tangents.

Anyway, as I like to see it, $dy$ means that you change the y coordinate by a very small amount. As a result the x coordinate also changes by an amount $dx$.
Their ratio happens to be the derivative, when we take those changes to the limit of 0.

But if you want, you can also look at it as x being a function of y, the inverse function.
The symbol $y$ is actually the function $y(x)$. If that function is invertible, then its inverse can be written as $x(y)=y^{-1}(y)$.
The derivative of $y(x)$ is $\frac{dy}{dx}$. The derivative of its inverse is $\frac{dx}{dy}$.

#### alane1994

##### Active member
Thank you very much for your help!