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If y=axn+2+bxn+1y = ax^{n+2} + \dfrac{b}{x^{n+1}}, then x2d2ydx2x^2\dfrac{d^2y}{dx^2} is equal to

Solution

✅ Correct Option: 2

Differentiating twice, d2ydx2=a(n+2)(n+1)xn+b(n+1)(n+2)x−(n+3)\dfrac{d^2y}{dx^2} = a(n+2)(n+1)x^n + b(n+1)(n+2)x^{-(n+3)}.

Multiplying by x2x^2 gives (n+1)(n+2)[axn+2+bx−(n+1)](n+1)(n+2)\left[ax^{n+2}+bx^{-(n+1)}\right].

The bracket is exactly yy, so x2d2ydx2=(n+1)(n+2)yx^2\dfrac{d^2y}{dx^2} = (n+1)(n+2)y.

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