If y = a{x^{n + 1}} + b{x^{ - n}}, then {x^2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $y = a{x^{n + 1}} + b{x^{ - n}}$Differentiating with respect to $x$:$\frac{dy}{dx} = a(n + 1){x^n} - bn{x^{ - n - 1}}$Differentiating again

Vedclass · Accepted Answer

$n(n + 1)y$

Vedclass · Answer

(B) Given: $y = a{x^{n + 1}} + b{x^{ - n}}$Differentiating with respect to $x$:$\frac{dy}{dx} = a(n + 1){x^n} - bn{x^{ - n - 1}}$Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = a(n + 1)n{x^{n - 1}} - bn(-n - 1){x^{ - n - 2}}$$\frac{d^2y}{dx^2} = an(n + 1){x^{n - 1}} + bn(n + 1){x^{ - n - 2}}$Now,multiply by ${x^2}$:${x^2}\frac{d^2y}{dx^2} = {x^2}[an(n + 1){x^{n - 1}} + bn(n + 1){x^{ - n - 2}}]$${x^2}\frac{d^2y}{dx^2} = an(n + 1){x^{n + 1}} + bn(n + 1){x^{ - n}}$${x^2}\frac{d^2y}{dx^2} = n(n + 1)[a{x^{n + 1}} + b{x^{ - n}}]$Since $y = a{x^{n + 1}} + b{x^{ - n}}$,we get:${x^2}\frac{d^2y}{dx^2} = n(n + 1)y$

If $y = a{x^{n + 1}} + b{x^{ - n}},$ then ${x^2}\frac{{{d^2}y}}{{d{x^2}}}$ is equal to:

Similar Questions

Let $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$,$x \in [0, \pi] - \{\frac{\pi}{4}\}$. Then $f(\frac{7\pi}{12}) f''(\frac{7\pi}{12})$ is equal to

If $f(x) = 10 \cos x + (13 + 2x) \sin x$, then $f''(x) + f(x)$ is equal to

If $y=f(x)$ is a thrice differentiable function and a bijection,then $\frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=$

If $x = A \cos 4t + B \sin 4t$,then $\frac{d^2x}{dt^2} = $

If $x^m + y^m = k$ where $(m \neq 1)$ and $\frac{d^2y}{dx^2} = \frac{ax^b}{y^c}$ such that $a + b + c = 0$, then the value of $k$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module