Solve the initial-value problem dy/dx = 2xy with y(0) = 3.

Step 1: Separate the variables.

The differential equation is:
dy/dx = 2xy

For y ≠ 0, divide by y and multiply by dx:

(1/y) dy = 2x dx

Step 2: Integrate both sides.

∫ (1/y) dy = ∫ 2x dx

This gives:
ln|y| = x^2 + C

Step 3: Solve for y.

Exponentiate both sides:

|y| = e^(x^2 + C)

Absorb the constant and possible sign into a new arbitrary constant A:

y = Ae^(x^2)

Step 4: Apply the initial condition.

We are given:
y(0) = 3

Substitute x = 0 and y = 3:

3 = Ae^(0^2)

Since e^0 = 1:
3 = A

Step 5: Write the particular solution.

Substitute A = 3:

y = 3e^(x^2)

Step 6: Verify the initial condition and differential equation.

At x = 0:
y(0) = 3e^0 = 3

Differentiate:
y' = 3e^(x^2)(2x) = 6xe^(x^2)

Also:
2xy = 2x(3e^(x^2)) = 6xe^(x^2)

The expressions agree.

Therefore, the solution to the initial-value problem is y = 3e^(x^2).