The particular solution of the differential equation $\frac{dy}{dx} = 8yx$ when $y = 1$ at $x = 0$
Answer & explanation
Correct answer: option 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
$y = e^{4x^2}$