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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
If  $x = \int\limits_{ - y}^y {\frac{{dt}}{{\sqrt {1 + 9{t^2}} }}\,and\,\frac{{{d^2}y}}{{d{x^2}}} = ky} $, then $k$ equals
  • A
    $9$
  • ✓
    $\frac{9}{4}$
  • C
    $\frac{9}{2}$
  • D
    $18$
✓

Answer

Correct option: B.
$\frac{9}{4}$
b
$\mathrm{x}=2 \int_{0}^{\mathrm{y}} \frac{\mathrm{dt}}{\sqrt{1+9 \mathrm{t}^{2}}} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{2}{\sqrt{1+9 \mathrm{y}^{2}}}$

$\frac{d y}{d x}=\frac{\sqrt{1+9 y^{2}}}{2}$

$\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \cdot \frac{1}{2} \frac{18 y}{\sqrt{1+9 y^{2}}} \cdot \frac{\sqrt{1+9 y^{2}}}{2}=\frac{9}{4} y$

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