Originally Posted by

**Diemo** So far so good, you have

$\displaystyle y=\frac{1}{\frac{1}{y_0}-t^2}$.

Now rearrange this a litlle by multiplying the top and the bottom by $\displaystyle y_0$ to get

$\displaystyle y=\frac{y_0}{1-y_0t^2}$.

Now your correct in saying that this exists provided the denominator is non-zero, or when $\displaystyle y_0t^2 \not = 1$. But note that $\displaystyle t^2$ is always greater than zero, hence when $\displaystyle y_0$ is negative, it is impossible for $\displaystyle t^2y_0$ to be positive, hence the equation exists for all $\displaystyle y_0<0, t$. For $\displaystyle y_0>0$ the denominator is equal to zero when $\displaystyle t=+-\frac{1}{\sqrt{y_0}}$.

That agree with your book?