If y=cot−1(tan(x3)+sec(x3)),π2<x3<3π2 then

y=cot−1(1+sin(x3)cos(x3))=cot−1(1+cos(π2−x3)sin(π2−x3)) = cot−1(2cos2(π4−x32)2sin(π4−x32)cos(π4−x32)) = cot−1(cot(π4−x32)) π2<x3<3π2⇒π4<x32<3π4 −3π4<−x32<−π4 −π2<π4−x32<0 (π4−x32)∈(−π2,0) cot−1(cot(π4−x32))=π+(π4−x32)=5π4−x32 y=5π4−x32 dydx=−3x22 d2ydx2=−3x y=5π4−x22(−y113) ⇒x2y116+5π4=y ⇒x2y11−6y+15π2=0

If y=cot−1(tan(x3)+sec(x3)),π2<x3<3π2 then

y=cot−1(1+sin(x3)cos(x3))=cot−1(1+cos(π2−x3)sin(π2−x3)) = cot−1(2cos2(π4−x32)2sin(π4−x32)cos(π4−x32)) = cot−1(cot(π4−x32)) π2<x3<3π2⇒π4<x32<3π4 −3π4<−x32<−π4 −π2<π4−x32<0 (π4−x32)∈(−π2,0) cot−1(cot(π4−x32))=π+(π4−x32)=5π4−x32 y=5π4−x32 dydx=−3x22 d2ydx2=−3x y=5π4−x22(−y113) ⇒x2y116+5π4=y ⇒x2y11−6y+15π2=0

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If $y = \cot^{- 1} (\tan (x^{3}) + \sec (x^{3})), \frac{π}{2} < x^{3} < \frac{3 π}{2}$ then

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$x^{2} y^{11} - 6 y + \frac{3 π}{2} = 0$

$x y^{11} + 3 y^{1} = 0$

$x^{2} y^{11} - 6 y + \frac{15 π}{2} = 0$

$x y^{11} - 6 y^{1} = 0$

answer is C.

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Detailed Solution

$y = \cot^{- 1} (\frac{1 + \sin (x^{3})}{\cos (x^{3})}) = \cot^{- 1} (\frac{1 + \cos (\frac{π}{2} - x^{3})}{\sin (\frac{π}{2} - x^{3})})$
= $\cot^{- 1} (\frac{2 \cos^{2} (\frac{π}{4} - \frac{x^{3}}{2})}{2 \sin (\frac{π}{4} - \frac{x^{3}}{2}) \cos (\frac{π}{4} - \frac{x^{3}}{2})})$
= $\cot^{- 1} (\cot (\frac{π}{4} - \frac{x^{3}}{2}))$
$\frac{π}{2} < x^{3} < \frac{3 π}{2} \Rightarrow \frac{π}{4} < \frac{x^{3}}{2} < \frac{3 π}{4}$
$- \frac{3 π}{4} < \frac{- x^{3}}{2} < \frac{- π}{4}$
$- \frac{π}{2} < \frac{π}{4} - \frac{x^{3}}{2} < 0$
$(\frac{π}{4} - \frac{x^{3}}{2}) \in (\frac{- π}{2}, 0)$
$\cot^{- 1} (\cot (\frac{π}{4} - \frac{x^{3}}{2})) = π + (\frac{π}{4} - \frac{x^{3}}{2}) = \frac{5 π}{4} - \frac{x^{3}}{2}$
$y = \frac{5 π}{4} - \frac{x^{3}}{2}$
$\frac{d y}{d x} = \frac{- 3 x^{2}}{2}$
$\frac{d^{2} y}{d x^{2}} = - 3 x$
$y = \frac{5 π}{4} - \frac{x^{2}}{2} (\frac{- y^{11}}{3})$
$\Rightarrow \frac{x^{2} y^{11}}{6} + \frac{5 π}{4} = y$
$\Rightarrow x^{2} y^{11} - 6 y + \frac{15 π}{2} = 0$