Let \(y = y(x)\) be the solution of the differential equation \(\displaystyle\frac{dy}{dx} = 2xy\) satisfying \(y(0) = 2\).
(A) The general solution of the differential equation is \(y = Ce^{x^2}\), C an arbitrary constant
(B) \(y(1) = 2e\)
(C) \(y(-1) = 2e^{-1}\)
(D) The value of \(\displaystyle\frac{dy}{dx}\) at \(x = 1\) is \(2e\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
The equation is variables separable:
\( \displaystyle \frac{dy}{y} = 2x\,dx \)
\(\log|y| = x^2 + C_1\)
\(y = Ce^{x^2}\)
Putting x = 0, y = 2 gives C = 2:
\(y = 2e^{x^2}\)
\( \displaystyle \frac{dy}{dx} = 2xy = 4x\,e^{x^2} \)
So the general solution is \(y = Ce^{x^2}\), the particular solution is \(y = 2e^{x^2}\), y(1) = y(−1) = 2e, and dy/dx at x = 1 is 4e.
Checking the statements
(A) general solution \(y = Ce^{x^2}\) — Correct. It comes from \(\log|y| = x^2 + C_1\).
(B) \(y(1) = 2e\) — Correct. \(2e^{1} = 2e\).
(C) \(y(-1) = 2e^{-1}\) — Incorrect. \((-1)^2 = 1\), so \(y(-1) = 2e\).
(D) \(\frac{dy}{dx} = 2e\) at x = 1 — Incorrect. \(2 \times 1 \times 2e = 4e\).