Hi again. Here's one.

(a) Without solving, explain why the intitial value problem

$\displaystyle \frac{dy}{dx}=\sqrt{y},$ $\displaystyle y(x_0)=y_0$

has no solution for $\displaystyle y_0<0$.

(b) Solve the initial value problem in part (a) for $\displaystyle y_0>0$ and find the largest interval $\displaystyle I$ on which the solution is defined.

My attempt:

(a) Because the only way to arrive at a radical through differentiation is to begin with one, and any expression involving a radical (even root) with nothing outside of the radical must always give a positive result over the reals.

(b) I get

$\displaystyle y=\frac{1}{4}(x+c)^2$.

If I set $\displaystyle y_0=\frac{1}{4}(x_0+c)^2$, I get $\displaystyle c=\pm2\sqrt{y}-x_0$.

But, I won't evfen attempt to find the interval of definition because the book doesn't include the plus or minus. They have $\displaystyle c=2\sqrt{y}-x_0$.

What's the deal here?