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$x{{{d^2}y} \over {d{x^2}}} = 2xy$
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