We are tasked with finding the particular solution to the differential equation $y' = 2xy$, which passes through the point $(0, \frac{1}{2})$, given that the general solution is $y = Ce^{x^2}$.

Steps to find the particular solution:

1. General Solution:
   We are provided with the general solution $y = Ce^{x^2}$, where $C$ is an arbitrary constant.

2. Use the Initial Condition:
   We are given that the particular solution passes through $(0, \frac{1}{2})$. This means that when $x = 0$, $y = \frac{1}{2}$.

   Substituting $x = 0$ and $y = \frac{1}{2}$ into the general solution: $\frac{1}{2} = C e^{0^2}$ Since $e^{0^2} = e^0 = 1$, we have: $\frac{1}{2} = C$

3. Find the Particular Solution:
   From the above, we find that $C = \frac{1}{2}$.

   Therefore, the particular solution is: $y = \frac{1}{2} e^{x^2}$

Final Answer:

The particular solution to the differential equation $y' = 2xy$ that passes through $(0, \frac{1}{2})$ is: $y = \frac{1}{2} e^{x^2}$

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Would you like a more detailed explanation or have further questions on this topic? Here are some related questions to explore:

1. How can we verify that $y = \frac{1}{2} e^{x^2}$ satisfies the differential equation $y' = 2xy$?
2. What is the method to solve first-order linear differential equations like this one?
3. How would the solution change if the initial condition were $(0, 1)$?
4. Can this general solution form $y = Ce^{x^2}$ be applied to other differential equations?
5. How does the method of separation of variables work in solving equations like $y' = 2xy$?

Tip: Always check your solution by substituting it back into the original differential equation to verify correctness!