trautlashley4649 trautlashley4649
  • 15-01-2020
  • Mathematics
contestada

Find the particular solution that satifies the differential equation and the initial condition. f"(x) = x2 f'(0) = 8, f(0) = 4

Respuesta :

akivieobukomena akivieobukomena
  • 16-01-2020

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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