What is the interval for the particular solution?

[faradayscat]

Homework Statement

dy/dx = (9x²)/(9y²-11) , y(1)=0

Homework Equations

The Attempt at a Solution

I've found the particular solution:

3y³-11y = 3x³-3

The solution is defined for -⅓√11<y<⅓√11,
What about the x values? Plugging the y values into the equation doesn't work of course..

 

[RUber]

It looks to me that your solution is defined everywhere...how did you conclude the interval above?

 

[RUber]

I see now...since you start with y=1, there is no way to continuously make it outside that interval -- where $9y^2 - 11 = 0$.
To solve for the domain of x, I would recommend finding the range (max, min) of the function of y over the interval of definition. This would give you a domain defined by:
$3x^3 - 3 \in \left[ \min_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\} , \max_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\}\right].$

 

[faradayscat]

I see now...since you start with y=1, there is no way to continuously make it outside that interval -- where $9y^2 - 11 = 0$.
To solve for the domain of x, I would recommend finding the range (max, min) of the function of y over the interval of definition. This would give you a domain defined by:
$3x^3 - 3 \in \left[ \min_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\} , \max_{y \in ( -\sqrt{11}/3, \sqrt{11}/3)} \{3y^3 -11\}\right].$

Yes I understand all of this, however my assignment wants me to show the interval of existence like this:

A < x < B

That is, to find the domain, not the range. How can I do that?

 

[faradayscat]

I don't think actual values are possible here? Only approximations of "x" for values greater by closest to -⅓√11 and smaller but closest to ⅓√11...

 

[RUber]

The function of y, $f(y) = 3y^3-11y$ has its own domain and range. The domain is $-\sqrt{11}/3 < y < \sqrt{11}/3.$ What is its range?
If $g(x)=3x^3 - 3$, then the range of f(y) must equal the range of g(x). Then you find the domain of x that restricts g to the range you have found.
For example if the minimum of f(y) is -10, you would say:
$g(x) > -10 \\ 3x^3 - 3 > -10 \\ x^3 > -7/3 \\ x > (-7/3)^{1/3}$