The particular solution of the differential equation \(\displaystyle\frac{dy}{dx} = 8yx\) when \(y = 1\) at \(x = 0\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(y = e^{4x^2}\)
The equation is in variables separable form:
\( \displaystyle \frac{dy}{y} = 8x\,dx \)
Integrating both sides:
\(\log|y| = 4x^2 + C\)
Putting \(y = 1\) at \(x = 0\):
\(\log 1 = 0 + C,\quad C = 0\)
So \(\log|y| = 4x^2\), and since y is positive (y = 1 at x = 0):
\(y = e^{4x^2}\)
Shortcut: Every option gives y = 1 at x = 0, so the initial condition alone does not decide. Differentiating Option (2):
\( \displaystyle \frac{dy}{dx} = 8x\,e^{4x^2} = 8xy \)
which is the given equation. Options (1), (3) and (4) give \(2xy\), \(16xy\) and \(y\) instead.