Find the particular solution that satifies the differential equation and the initial condition. f"(x) = x2 f'(0) = 8, f(0) = 4
1 answer:
Answer:
f(x) = x^4/12 + 8x + 4
Step-by-step explanation:
f"(x) = x^2
Integrate both sides with respect to x
f'(x) = ∫ x^2 dx
= (x^2+1)/2+1
= (x^3)/3 + C
f(0) = 8
Put X = 0
f'(0) = 0+ C
8 = 0 + C
C= 8
f'(x) = x^3/3 + 8
Integrate f(x) again with respect to x
f(x) = ∫ (x^3 / 3 ) +8 dx
= ∫ x^3 / 3 dx + ∫8dx
= x^(3+1) / 3(3+1) + 8x + D
= x^4/12 + 8x + D
f(0) = 4
Put X = 0
f(0) = 0 + 0 + D
4 = D
Therefore
f(x) = x^4 /12 + 8x + 4
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