MCQ
If $y = a + b{x^2}$;    $a$ , $b$ arbitrary constants, then
  • A
    ${{{d^2}y} \over {d{x^2}}} = 2xy$
  • $x{{{d^2}y} \over {d{x^2}}} = {{dy} \over {dx}}$
  • C
    $x{{{d^2}y} \over {d{x^2}}} - {{dy} \over {dx}} + y = 0$
  • D
    $x{{{d^2}y} \over {d{x^2}}} = 2xy$

Answer

Correct option: B.
$x{{{d^2}y} \over {d{x^2}}} = {{dy} \over {dx}}$
b
(b)$\frac{{dy}}{{dx}} = 2bx,\;\;\frac{{{d^2}y}}{{d{x^2}}} = 2b$==> $x\frac{{{d^2}y}}{{d{x^2}}} = 2bx = \frac{{dy}}{{dx}}$.

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