For the differential equation \(\displaystyle\frac{dy}{dx} = 2xy\), \(y \gt 0\), consider the following statements:
(A) Its order is 1 and its degree is 1.
(B) Its general solution is \(y = Ce^{x^2}\), where C is an arbitrary constant.
(C) Its particular solution satisfying \(y(0) = 2\) is \(y = e^{x^2} + 1\).
(D) Its solution curve through the point (0, 1) also passes through the point (1, e).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Only the first derivative appears, with power 1. Separating the variables:
\( \displaystyle \frac{dy}{y} = 2x\,dx \)
\(\log y = x^2 + C_1\)
\(y = e^{C_1}e^{x^2} = Ce^{x^2}\)
With y(0) = 2: \(C = 2\), so \(y = 2e^{x^2}\).
With y(0) = 1: \(C = 1\), so \(y = e^{x^2}\), and at x = 1, \(y = e\).
So the order and degree are 1 and 1, the general solution is \(y = Ce^{x^2}\), the solution with y(0) = 2 is \(y = 2e^{x^2}\), and the curve through (0, 1) is \(y = e^{x^2}\).
Checking the statements
(A) Its order is 1 and its degree is 1. — Correct. Only \(\frac{dy}{dx}\) appears, with power 1.
(B) Its general solution is \(y = Ce^{x^2}\), where C is an arbitrary constant. — Correct. \(\log y = x^2 + C_1\) gives \(y = Ce^{x^2}\).
(C) Its particular solution satisfying \(y(0) = 2\) is \(y = e^{x^2} + 1\). — Incorrect. It is \(y = 2e^{x^2}\); \(e^{x^2} + 1\) gives y = 2 at x = 0 but has slope \(2xe^{x^2}\), not \(2x(e^{x^2} + 1)\).
(D) Its solution curve through the point (0, 1) also passes through the point (1, e). — Correct. The curve is \(y = e^{x^2}\), and \(e^{1} = e\).