Question

If \(a\) and \(b\) are arbitrary constants, then the differential equation corresponding to the family of curves \[ ax^2+2hxy=1 \] is

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

Hint:

The order of the differential equation equals the number of arbitrary constants present in the family of curves.

Answer 1

The Correct Option is C

Step 1: Count the constants.
The family $ax^2+2hxy=1$ has two free constants $a$ and $h$, so we differentiate twice to remove them.
Step 2: Differentiate once.
\[ 2ax+2h\left(y+x\frac{dy}{dx}\right)=0\Rightarrow ax+hy+hx\frac{dy}{dx}=0.\quad(1) \]
Step 3: Differentiate again.
\[ a+h\frac{dy}{dx}+h\left(\frac{dy}{dx}+x\frac{d^2y}{dx^2}\right)=0\Rightarrow a+2h\frac{dy}{dx}+hx\frac{d^2y}{dx^2}=0.\quad(2) \]
Step 4: Use (1) to express $ax$.
From (1), $ax=-hy-hx\dfrac{dy}{dx}$, so $a=-h\dfrac{y}{x}-h\dfrac{dy}{dx}$.
Step 5: Substitute into (2).
\[ -h\frac{y}{x}-h\frac{dy}{dx}+2h\frac{dy}{dx}+hx\frac{d^2y}{dx^2}=0. \] Divide by $h$ and tidy: $-\dfrac{y}{x}+\dfrac{dy}{dx}+x\dfrac{d^2y}{dx^2}=0$.
Step 6: Multiply by $x$.
\[ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0. \] \[ \boxed{x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=0} \]