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

Concept: To form a differential equation from a family of curves containing arbitrary constants, differentiate sufficiently many times and eliminate the constants. Since there are two arbitrary constants, we differentiate twice.

Step 1: Write the given family. \[ ax^2+2hxy=1 \] Differentiate with respect to \(x\): \[ 2ax+2h\left(y+x\frac{dy}{dx}\right)=0 \] \[ ax+h\left(y+x\frac{dy}{dx}\right)=0 \] \[ ax+hy+hx\frac{dy}{dx}=0 \quad \cdots (1) \]

Step 2: Differentiate again. Differentiating equation (1): \[ a+h\left(\frac{dy}{dx}\right) +h\left(\frac{dy}{dx}+x\frac{d^2y}{dx^2}\right)=0 \] \[ a+2h\frac{dy}{dx} +hx\frac{d^2y}{dx^2}=0 \quad \cdots (2) \]

Step 3: Eliminate constants. From the original equation, \[ ax^2+2hxy=1 \] From equation (1), \[ ax=-hy-hx\frac{dy}{dx} \] Substitute into equation (2): \[ -h\frac{y}{x} -h\frac{dy}{dx} +2h\frac{dy}{dx} +hx\frac{d^2y}{dx^2}=0 \] \[ -h\frac{y}{x} +h\frac{dy}{dx} +hx\frac{d^2y}{dx^2}=0 \] Dividing by \(h\), \[ -\frac{y}{x} +\frac{dy}{dx} +x\frac{d^2y}{dx^2}=0 \] Multiplying by \(x\), \[ x^2\frac{d^2y}{dx^2} +x\frac{dy}{dx} -y=0 \] Hence, the required differential equation is \[ \boxed{ x^2\frac{d^2y}{dx^2} +x\frac{dy}{dx} -y=0 } \]