Solveeit Logo
Login

Question

Question: If \(y = \left( x + \sqrt{1 + x^{2}} \right)^{n},\) then \((1 + x^{2})\frac{d^{2}y}{dx^{2}} + x\frac...

If y=(x+1+x2)n,y = \left( x + \sqrt{1 + x^{2}} \right)^{n}, then (1+x2)d2ydx2+xdydx(1 + x^{2})\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} is

A

n2yn^{2}y

B

−n2y- n^{2}y

C

−y- y

D

2x2y2x^{2}y

Answer

n2yn^{2}y

Explanation

Solution

y=(x+1+x2)ny = (x + \sqrt{1 + x^{2}})^{n}⇒ dydx=n(x+1+x2)n−1(1+x1+x2)\frac{dy}{dx} = n(x + \sqrt{1 + x^{2}})^{n - 1}\left( 1 + \frac{x}{\sqrt{1 + x^{2}}} \right)

⇒ dydx=n(x+1+x2)n1+x2\frac{dy}{dx} = \frac{n(x + \sqrt{1 + x^{2}})^{n}}{\sqrt{1 + x^{2}}} ⇒ (1+x2)dydx=n(x+1+x2)n(\sqrt{1 + x^{2}})\frac{dy}{dx} = n\left( x + \sqrt{1 + x^{2}} \right)^{n}

⇒d2ydx2.1+x2+dydx(x1+x2)\frac{d^{2}y}{dx^{2}}.\sqrt{1 + x^{2}} + \frac{dy}{dx}\left( \frac{x}{\sqrt{1 + x^{2}}} \right)

=n2(x+1+x2)n−1(1+x1+x2)= n^{2}\left( x + \sqrt{1 + x^{2}} \right)^{n - 1}\left( 1 + \frac{x}{\sqrt{1 + x^{2}}} \right)

⇒ (1+x2).d2ydx2+x.dydx=n2(x+1+x2)n(1 + x^{2}).\frac{d^{2}y}{dx^{2}} + x.\frac{dy}{dx} = n^{2}(x + \sqrt{1 + x^{2}})^{n}

⇒ (1+x2)d2ydx2+x.dydx=n2y(1 + x^{2})\frac{d^{2}y}{dx^{2}} + x.\frac{dy}{dx} = n^{2}y