- #1

amcavoy

- 665

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I have the following problem:

[tex]y'=\frac{3x^{2}}{3y^{2}-4};\quad y(1)=0[/tex]

I solved this, and came up with [itex]y^{3}-4y-x^{3}=-1[/itex]

My problem lies in finding the interval of definition. From the original DE, it is clear that [itex]|y|\neq \frac{2\sqrt{3}}{3}[/itex]. Then, after plugging that in for y and solving for what x

[tex]\left|x^{3}-1\right|\neq \frac{16\sqrt{3}}{9}[/tex]

...however, my reasoning doesn't seem to work here. Apparently, the answer is:

[tex]\left|x^{3}-1\right|<\frac{16\sqrt{3}}{3}[/tex]

Can someone help me out here?

Thanks a lot.

[tex]y'=\frac{3x^{2}}{3y^{2}-4};\quad y(1)=0[/tex]

I solved this, and came up with [itex]y^{3}-4y-x^{3}=-1[/itex]

My problem lies in finding the interval of definition. From the original DE, it is clear that [itex]|y|\neq \frac{2\sqrt{3}}{3}[/itex]. Then, after plugging that in for y and solving for what x

**can't**be, I come up with:[tex]\left|x^{3}-1\right|\neq \frac{16\sqrt{3}}{9}[/tex]

...however, my reasoning doesn't seem to work here. Apparently, the answer is:

[tex]\left|x^{3}-1\right|<\frac{16\sqrt{3}}{3}[/tex]

Can someone help me out here?

Thanks a lot.

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