The differential equation of the family of curves \(y = ax^2 + bx\), where a and b are arbitrary constants, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x^2\displaystyle\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0\)
There are two arbitrary constants, so the equation is of order 2. Differentiating twice:
\( \displaystyle \frac{dy}{dx} = 2ax + b \)
\( \displaystyle \frac{d^2y}{dx^2} = 2a \)
Eliminating a and b: from these,
\( \displaystyle a = \frac{1}{2}\frac{d^2y}{dx^2} \)
\( \displaystyle b = \frac{dy}{dx} - x\frac{d^2y}{dx^2} \)
Substituting in \(y = ax^2 + bx\):
\( \displaystyle y = \frac{x^2}{2}\frac{d^2y}{dx^2} + x\frac{dy}{dx} - x^2\frac{d^2y}{dx^2} \)
\( \displaystyle y = x\frac{dy}{dx} - \frac{x^2}{2}\frac{d^2y}{dx^2} \)
Multiplying by 2 and rearranging:
\( \displaystyle x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0 \)
Why the other options are incorrect
Option (1) \(x^2\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - 2y = 0\) — Incorrect. For \(y = x^2\) (a = 1, b = 0) the left side is \(2x^2 + 4x^2 - 2x^2 = 4x^2 \ne 0\).
Option (3) \(x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} - 2y = 0\) — Incorrect. For \(y = x\) (a = 0, b = 1) the left side is \(0 - 2x - 2x = -4x \ne 0\).
Option (4) \(x\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0\) — Incorrect. For \(y = x\) the left side is \(0 - 2 + 2x\), which is not zero for every x.