Consider the family of curves \(y = ax^2\), where a is an arbitrary constant.
(A) The differential equation of this family is of order 1.
(B) The differential equation of this family is \(x\displaystyle\frac{dy}{dx} = 2y\).
(C) \(y = 5x^2\) is a particular solution of this differential equation.
(D) \(y = x^2 + 1\) is a solution of this differential equation.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
The family has one arbitrary constant, so its differential equation is of order 1.
\( \displaystyle \frac{dy}{dx} = 2ax \)
Putting \(a = \frac{y}{x^2}\):
\( \displaystyle \frac{dy}{dx} = \frac{2y}{x} \)
\( \displaystyle x\frac{dy}{dx} = 2y \)
A particular solution is a member of the family with a fixed value of a.
So the equation is \(x\frac{dy}{dx} = 2y\), of order 1, and its solutions are the curves \(y = ax^2\).
Checking the statements
(A) The differential equation is of order 1 — Correct. There is only one constant to eliminate.
(B) The differential equation is \(x\frac{dy}{dx} = 2y\) — Correct. It comes from \(\frac{dy}{dx} = 2ax\) with \(a = \frac{y}{x^2}\).
(C) \(y = 5x^2\) is a particular solution — Correct. It is the member with a = 5.
(D) \(y = x^2 + 1\) is a solution — Incorrect. Left side \(x \cdot 2x = 2x^2\), right side \(2x^2 + 2\).