Q72 / 125 · BITSAT 2012 · Mathematics

The differential equation whose solution is $A x^{2}+B y^{2}=1$ where $A$ and $B$ are arbitrary constants is of

  • A
    second order and second degree
  • B
    first order and second degree
  • C
    first order and first degree
  • Correct answer
    D
    second order and first degree

Solution

$A x^{2}+B y^{2}=1$ ...(1) $A x+B y \frac{d y}{d x}=0$ ...(2) $A+B y \frac{d^{2} y}{d x^{2}}+B\left(\frac{d y}{d x}\right)^{2}=0$ ...(3) $x\left\{-B y \frac{d^{2} y}{d x^{2}}-B\left(\frac{d y}{d x}\right)^{2}\right\}+B y \frac{d y}{d x}=0$ Dividing both sides by $-\mathrm{B}$, we get $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$ Which is a DE of order 2 and degree 1

Hence, the correct answer is option: 'D'.

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