I'm working on a small problem.
Find the solution of the initial value problem $y' = \frac{2x}{1 + 2y}$, $y(2) = 0$ in explicit form and obtain the following solution
$$y = \frac{1}{2} \left(-1 + \sqrt{4x^2-15}\right)$$
The question I'm having is on the domain of validity of the solution. I obtain the interval $\left(\frac{\sqrt{15}}{2},\infty\right) \bigcup \left(-\infty,-\frac{\sqrt{15}}{2}\right)$. However, I am told that only the latter interval should be included since the solution should be continuous. May I ask if this is indeed the case, and why so? Thank you.
The interval of validity for this IVP is just the interval $(\frac{\sqrt{15}}{2},\infty ) $because it contains the initial condition that $x = 2$. The other interval of validity wouldn't work because your initial condition isn't in that set.
If this were a regular ODE with no specified initial condition, then what you have right now would be the correct intervals of validity.
Hope that helps!