Question

Y = 10 (in sec x cube minus 10 x cube bracket close pi by 2 is greater than x cube greater than 3/2

• A. x \cdot y'' + 2y' = 0
• B. x^2 \cdot y'' - 6y + 3\pi/12 = 0
• C. x^2 \cdot y'' - 6y + 3\pi = 0
• D. x \cdot y'' - 4y' = 0

Answer 1

None of the options is correct

Answer 2

We start with

$$y=\tan^{-1}\Bigl(\sec x^3-\tan x^3\Bigr),\quad \text{with} \quad \frac{\pi}{2}<x^3<\frac{3\pi}{2}.$$

A standard trigonometric result shows that

$$\sec x^3-\tan x^3=\tan\Bigl(\frac{\pi}{4}-\frac{x^3}{2}\Bigr).$$

Thus,

$$y=\tan^{-1}\Bigl(\tan\Bigl(\frac{\pi}{4}-\frac{x^3}{2}\Bigr)\Bigr) =\frac{\pi}{4}-\frac{x^3}{2},$$

since the angle $\frac{\pi}{4}-\frac{x^3}{2}$ lies in the principal range $(-\pi/2,\pi/2)$.

Differentiate:

$$y'=-\frac{3}{2}x^2,\quad y''=-3x.$$

It can be verified that none of the candidate differential equations provided (after substitution) lead to an identity. For example, testing one option:

$$x\,y''+2y' = x(-3x)+2\Bigl(-\frac{3}{2}x^2\Bigr) =-3x^2-3x^2=-6x^2\neq0.$$

Thus, none of the given options is correct.

---

Explanation (minimal):
Use the identity $\sec\theta-\tan\theta=\tan(\pi/4-\theta/2)$ with $\theta=x^3$ to get $y=\pi/4-\frac{x^3}{2}$. Then differentiate to obtain $y'=-\frac{3}{2}x^2$ and $y''=-3x$. None of the given differential equations are identically satisfied by these expressions.