The particular solution of the differential equation $\frac{dy}{dx} = 8yx$ when $y = 1$ at $x = 0$

$y = e^{4x^2}$

The correct answer is Option (2) → $y = e^{4x^2}$

Given differential equation:

$\frac{dy}{dx} = 8xy$

Separate variables:

$\frac{dy}{y} = 8x\,dx$

Integrate both sides:

$\int \frac{1}{y}\,dy = \int 8x\,dx$

$\ln|y| = 4x^2 + C$

Take exponential on both sides:

$y = e^{4x^2 + C} = Ae^{4x^2}$, where $A = e^C$

Use initial condition $y = 1$ when $x = 0$:

$1 = A e^{0} \Rightarrow A = 1$

Therefore, the particular solution is

$y = e^{4x^2}$