First Order Separable Differential Equation IVP Solution based on cases Consider the initial value problem
$$\ 4yy' = x, y(x_0)=1$$

a) Find the general solution to the ODE, solving for $y(x)$ explicitly.

b) Find the solutions to the initial value problem. What is the interval of existence? Note that you will need to separate your answer into cases, depending on $x_0$.

c) Sketch some solution curves in the $(x,y)$ plane, choosing at least one $x_0$ for each of the cases you identified in (b).

So far, I have solved part a), but I am unsure where to go from here. I plugged in $y=1$ and $x=x_0$ to solve for C, but I can't figure out what the interval of existence is/what the cases would be.
Thanks!
Note that

$$ y' = \frac{x}{4y} $$

is undefined at $y=0$. Thus, you cannot have a solution that contains $y=0$. Any solution must be either $y&lt;0$, or $y&gt;0$.

Since the given initial condition is $y(x_0)=1$, the solution must be $y &gt; 0$, i.e.

$$ y(x) = \frac{\sqrt{x^2-x_0^2 + 1}}{2} $$

You can then find the domain of this function depending on $x_0$
$4 \times y \times \frac{dy}{dx} =x$

$4 \times y \times dy =x \times dx$

Integrating both sides, we get

$4 \times \frac{y^2}{2}=\frac{x^2}{2} + c$