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Verified Answer :

syed514

d/dx (2 x^2 y + y = 2x + 13)

4xy + 2x^2 y' + y' = 2

4xy + y'(2x^2 + 1) = 2

y' = (2- 4xy)/(2x^2 +1)

ow we can use this in a linear equation for a slope

Ty = -5x/8 +5(3)/8 +8/8
= -5x/8 +(15+8)/8
= -5x/8 +23/8

this will gives us an approximation at x=2.8 now

vargasjerry

ANSWER:

y ≈ (-10/19)(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

STEP-BY-STEP EXPLANATION:

y = (2x+13)/(2x^2+1)


y' = ((2x^2+1)(2) - (2x+13)(4x)) / (2x^2+1)^2


at x=3, this is


y'(3) = (192 - 1912)/19^2 = -10/19


So, your linearization is


y ≈ (-10/19)(x-3) + 1


At x=2.8, this is


y ≈ (-10/19)(2.8 - 3) + 1 = 1 2/19 ≈ 1.105

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