The differential equation representing the family of curves $y = Ax+\frac{B}{x};x≠0$, where A and B are arbitrary constants, is given by
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0$
Given family of curves: $y = Ax + \frac{B}{x}$, $x \neq 0$
Differentiate w.r.t $x$:
$\frac{dy}{dx} = A - \frac{B}{x^2}$
Differentiate again:
$\frac{d^2y}{dx^2} = \frac{2B}{x^3}$
Multiply first derivative by $x$: $x\frac{dy}{dx} = xA - \frac{B}{x} = Ax - \frac{B}{x}$
Multiply second derivative by $x^2$: $x^2\frac{d^2y}{dx^2} = \frac{2B}{x^3} \cdot x^2 = \frac{2B}{x}$
Add $x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2} - y$:
$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = \frac{2B}{x} + (Ax - \frac{B}{x}) - (Ax + \frac{B}{x}) = 0$
Thus, the differential equation is:
$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - y = 0$