Solve the following initial value problems and find the corresponding solution curves :
`y'=2xy, y(0)=1`.

Answer

To solve the initial value problem given by the differential equation \( y' = 2xy \) with the initial condition \( y(0) = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the differential equation in a more manageable form:
\[ \frac{dy}{dx} = 2xy \]

### Step 2: Separate the variables
Next, we separate the variables \( y \) and \( x \):
\[ \frac{1}{y} dy = 2x dx \]

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{1}{y} dy = \int 2x dx \]
The left side integrates to \( \ln |y| \) and the right side integrates to \( x^2 + C \):
\[ \ln |y| = x^2 + C \]

### Step 4: Exponentiate to solve for \( y \)
To solve for \( y \), we exponentiate both sides:
\[ |y| = e^{x^2 + C} = e^{x^2} e^C \]
Let \( k = e^C \), where \( k \) is a positive constant:
\[ y = ke^{x^2} \]

### Step 5: Apply the initial condition
Now we apply the initial condition \( y(0) = 1 \):
\[ 1 = ke^{0^2} \implies 1 = k \cdot 1 \implies k = 1 \]
Thus, we have:
\[ y = e^{x^2} \]

### Final Solution
The solution to the initial value problem is:
\[ y = e^{x^2} \]