The correct option is D y _ 2 2 y _ 1 + y _ 0 Now, Δ y _ 0 = y _ 1 y _ 0 ⇒Δ #160;^ 2 y _ 0 =Δ( y _ 1 y _ 0 ) #160; #160; #160; ...

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Standard XII

Mathematics

Existence of Limit

Δ 2 y 0 is eq...

Question

$Δ^{2} y_{0}$ is equal to

2 y_{2} - 2 y_{1} - y_{0}

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y_{2} - 2 y_{1} - y_{0}

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2 y_{2} - 2 y_{1} + y_{0}

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y_{2} - 2 y_{1} + y_{0}

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Solution

The correct option is D $y_{2} - 2 y_{1} + y_{0}$
Now, $Δ y_{0} = y_{1} - y_{0}$
$\Rightarrow Δ^{2} y_{0} = Δ (y_{1} - y_{0})$
$= Δ y_{1} - Δ y_{0}$
$= y_{2} - y_{1} - y_{1} + y_{0}$
$= y_{2} - 2 y_{1} + y_{0}$

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Similar questions

Q. In trapezoidal rule

\int_{a}^{b} f (x) d x = \frac{h}{2} [y_{0} + 2 (y_{1} + y_{2} + . . . . . y_{a - 1}) + y_{a}

. Write the value of

h

Q. Assertion(

A

): lf

e = 2.72, e^{2} = 7.39

e^{3} = 20.09, e^{4} = 54.6

then

\int_{0}^{4} e^{x} d x

by Simpson's rule is

53.873

Reason(R): Simpson's rule is

\int_{a}^{b} y d x = \frac{h}{2}

[(y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots . . + y_{n - 1})]

Q. Write the following transformation in matrix form

x_{1} = \frac{\sqrt{3}}{2} y_{1} + \frac{1}{2} y_{2}; x_{2} = - \frac{1}{2} y_{1} + \frac{\sqrt{3}}{2} y_{2}

.
Hence find the transformation in matrix form which expresses

y_{1}, y_{2}

in terms of

x_{1}, x_{2}

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

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

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

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

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

(viii)

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}

when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)

2 (y + 3) - x y \frac{d y}{d x} = 0

, y(1) = −2 [NCERT EXEMPLAR]

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Δ2y0 is equal to

The correct option is D y2−2y1+y0Now, Δy0=y1−y0⇒Δ2y0=Δ(y1−y0) =Δy1−Δy0 =y2−y1−y1+y0 =y2−2y1+y0

$Δ^{2} y_{0}$ is equal to

The correct option is D $y_{2} - 2 y_{1} + y_{0}$
Now, $Δ y_{0} = y_{1} - y_{0}$
$\Rightarrow Δ^{2} y_{0} = Δ (y_{1} - y_{0})$
$= Δ y_{1} - Δ y_{0}$
$= y_{2} - y_{1} - y_{1} + y_{0}$
$= y_{2} - 2 y_{1} + y_{0}$