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Mathematics Question on Trigonometric Equations

y=tan−1(sec  x3−tan  x3),π2<x3<3π2,y=tan^{-1}(sec \;x^3 - tan\; x^3), \frac{π}{2} < x^3< \frac{3π}{2}, then

A

xy′′+2y′=0xy′′ + 2y′ = 0

B

x2y′′−6y+3π2x^2y''-6y+\frac{3π}{2}

C

x2y″–6y+3π=0x^2y″ – 6y + 3π = 0

D

xy″–4y′=0xy″ – 4y′ = 0

Answer

x2y′′−6y+3π2x^2y''-6y+\frac{3π}{2}

Explanation

Solution

x3=θ⇒θ2∈(π4,3π4)x^3 = θ ⇒ \frac{θ}{2} ∈\bigg(\frac{π}{4}, \frac{3π}{4}\bigg)
∴y=tan–1(secθ–tanθ)∴ y = tan^{–1} (secθ – tanθ)

= tan−1(1−sinθcosθ)tan^{−1}(\frac{1−sinθ}{cos θ} )

∴y=π4−θ2.∴y=\frac{π}{4}−\frac{θ}{2}.

y=π4−x32y=\frac{π}{4}−\frac{x^3}{2}

∴y′=−3x22∴y′=\frac{−3x^2}{2}

y′′=–3xy'' = – 3x

∴x2y′′−6y+3π2=0∴ x^2y''-6y+\frac{3π}{2}=0

Hence, the correct option is (B): x2y′′−6y+3π2x^2y''-6y+\frac{3π}{2}