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Sagot :
Answer:
Start with equation,
y''(x) = f(x) = 6y/(10x-x²) about point x₀ = 5
Then,
y(x) = a₀ + a₁(x-5) + a₂(x-5)²+ a₃(x-5)³ + a₄(x-5)⁴...….
by inspection,
y(5) = a₀ and y'(5) = a₁
then,
y''(5) = 6a₀/(50-25) = 6a₀/25
y'''(x) = 6y'/(10x-x2) - 6y(10-2x)/(10x-x2)2
then,
y'''(5) = 6a₁/25 - 0 = 6a₁/25
y''''(x) = 6y''/(10x-x2) -12y'(10-2x)/(10x-x2)² + 12y/(10x-x2)² + 12y(10-2x)2/(10x-x2)³
then,
y''''(5) = (6/25)(6a₀/25) + 0 + 12a₀/625 + 0
= 48a₀/625
So,
y(x) = f(x) = f(5) + f'(5)(x-5) + f''(5)(x-5)2/2! + f'''(5)(x-5)3/3! + f''''(5)(x-5)4/4! …
or
y(x) = f(x) = a₀ + a₁(x-5) + (6ao/25)(x-5)2/2! + (6a1/25)(x-5)3/3! + (48a₀/625)(x-5)4/4!...
or
y(x) = f(x) = a₀ + a₁(x-5) + (3a₀/25)(x-5)²+ (a₁/25)(x-5)³+ (2a₀/625)(x-5)⁴.....
Learn more about equation here-
https://brainly.com/question/3391581
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