math

find the particular solution that satisfies the differential equation and the initial condition
f’’(x)=x^2 , f’(0)=4, f(0)=8

f(x)=

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  1. idk sorry

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  2. f" = x^2
    f' = 1/3 x^3 + C
    f'(0) = 4 ==> C=4, so
    f' = 1/3 x^3 + 4
    f = 1/12 x^4 + 4x + C
    f(0)=8 ==> C=8, so
    f(x) = 1/12 x^4 + 4x + 8

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    posted by oobleck

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