Question

If $y = tan^{-1}(secx^3 - tanx^3)$, $\pi/2 < x^3 < 3\pi/2$

• A. $x.y''' + 2y' = 0$
• B. $x^2.y'' - 6y + \frac{3\pi}{12} = 0$
• C. $x^2.y'' - 6y + 3\pi = 0$
• D. $x.y''' - 4y' = 0.$

Answer 1

None of the given options is correct.

Answer 2

We wish to “simplify”

$y = tan^{-1} (sec x^3 – tan x^3)$.

A standard trick is to write the expression in “sine‐cosine form”. First note that

$sec θ – tan θ = (1 – sin θ)/cos θ$.

Now with the substitution

$θ = x^3$ (so that $x^3$ lies between $π/2$ and $3π/2$)

write

$tan y = sec x^3 – tan x^3 = (1 – sin x^3)/cos x^3$.

A standard half–angle manipulation shows that

$(1 – sin x^3)/cos x^3 = (1 – t)/(1 + t)$ where $t = tan (x^3/2)$.

But one may verify that

$tan(π/4 – x^3/2) = (1 – tan(x^3/2))/(1 + tan(x^3/2)) = (1–t)/(1+t)$.

Thus

$tan y = tan (π/4 – x^3/2)$

so that (by the range of arctan)

$y = π/4 – x^3/2$.

This is an “explicit” expression for y. Being a polynomial (really a linear function of $x^3$) it has third derivative equal to the constant

$y''' = –3$.

Thus y satisfies

$y''' + 3 = 0$.

None of the options given in part (2) is equivalent to $y''' + 3 = 0$.