If y-a xn-1-b x-n- then x 2 d 2 y d x 2-a nn-1 yb nn-1 yc n yd n2 y

(b) n(n+1)y Here, y=axn+1+bx n #8658;dydx=an+1xn bn #160;x n 1 #8658;d2ydx2=ann+1xn 1+bnn+1x n 2 #8756; #160;x2d2ydx2=x2ann+1xn 1+bnn+1x n 2 #160; ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

If y=a xn+1+b...

Question

If y = axⁿ⁺¹ + bx⁻ⁿ, then $x^{2} \frac{d^{2} y}{d x^{2}} =$

(a) n (n − 1)y
(b) n (n + 1)y
(c) ny
(d) n²y

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Solution

(b) n(n+1)y

Here,
$y = a x^{n + 1} + b x^{- n} \Rightarrow \frac{d y}{d x} = a (n + 1) x^{n} - b n x^{- n - 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = a n (n + 1) x^{n - 1} + b n (n + 1) x^{- n - 2} ∴ x^{2} \frac{d^{2} y}{d x^{2}} = x^{2} \{a n (n + 1) x^{n - 1} + b n (n + 1) x^{- n - 2}\} = n (n + 1) (a x^{n + 1} + b x^{- n}) = n (n + 1) y$

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Similar questions

Q. If y = a xⁿ^{+ 1} + bx⁻ⁿ and

x^{2} \frac{d^{2} y}{d x^{2}} = λ y

, then write the value of λ.

If y = x^{n} \{a \cos (\log x) + b \sin (\log x)\}, prove that x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0 .

Disclaimer: There is a misprint in the question. It must be

x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0

instead of

x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) y = 0

Q. If

y = a x^{n + 1} + b x^{- n}

then

x^{2} y^{''} =

lim n \to \infty ((\frac{n}{n})^{n} + (\frac{n - 1}{n})^{n} + . . + ({\frac{1}{n}}^{n}))

equals

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

Find $a$

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If y = axn+1 + bx−n, then x2d2ydx2= (a) n (n − 1)y (b) n (n + 1)y (c) ny (d) n2y

(b) n(n+1)y Here, y=axn+1+bx-n⇒dydx=an+1xn-bn x-n-1⇒d2ydx2=ann+1xn-1+bnn+1x-n-2∴ x2d2ydx2=x2ann+1xn-1+bnn+1x-n-2 =nn+1axn+1+b x-n =nn+1y

If y = axⁿ⁺¹ + bx⁻ⁿ, then $x^{2} \frac{d^{2} y}{d x^{2}} =$

(a) n (n − 1)y
(b) n (n + 1)y
(c) ny
(d) n²y