Question

Elimination of arbitrary constants \(A\) and \(B\) from \(y=\frac{A}{x}+B\), \(x>0\) leads to the differential equation

• A. $x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=0$
• B. $x^{2}\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=0$
• C. $x^{2}\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}=0$
• D. $x\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}=0$
• E. $x\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}=0$

Hint:

Calculus Tip: The number of essential arbitrary constants in an equation always dictates the order of the differential equation required to represent it. 2 constants = 2nd order differential equation!

Answer 1

The Correct Option is A

Concept:
To form a differential equation from a given family of curves with $n$ arbitrary constants, we must differentiate the equation exactly $n$ times. Since there are two constants ($A$ and $B$), we will find the first and second derivatives and algebraically eliminate $A$ and $B$.

Step 1: Write the initial equation.
Given: $$y = Ax^{-1} + B$$

Step 2: Find the first derivative.
Differentiate with respect to $x$. The constant $B$ becomes 0: $$\frac{dy}{dx} = -Ax^{-2}$$

Step 3: Find the second derivative.
Differentiate again with respect to $x$: $$\frac{d^2y}{dx^2} = -A(-2x^{-3}) = 2Ax^{-3}$$ $$\frac{d^2y}{dx^2} = \frac{2A}{x^3}$$

Step 4: Eliminate the constant A.
From the first derivative equation, isolate $A$: $$\frac{dy}{dx} = -\frac{A}{x^2} \implies A = -x^2 \frac{dy}{dx}$$ Substitute this expression for $A$ into the second derivative equation: $$\frac{d^2y}{dx^2} = \frac{2\left(-x^2 \frac{dy}{dx}\right)}{x^3}$$

Step 5: Simplify into the final standard form.
Cancel $x^2$ from the numerator and denominator: $$\frac{d^2y}{dx^2} = -\frac{2}{x} \frac{dy}{dx}$$ Multiply both sides by $x$ to clear the fraction: $$x\frac{d^2y}{dx^2} = -2\frac{dy}{dx}$$ Move all terms to one side to set the equation to zero: $$x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0$$ Hence the correct answer is (A) $x\frac{d^{2y}{dx^{2}}+2\frac{dy}{dx}=0$}.