Solve the second order differential equation if f ''(x) = 3x^3 + x^2 + 4x + 5, f'(2) = 2/3 and f(3) = 10

To find f '(x), integrate f ''(x)

Therefore f '(x) = (3/4)x^4 + (1/3)x^3 + 2x^2 + 5x + C

Now substitute 2 for x and 2/3 for f ' (x)

2/3 = (3/4)(2^4) + (1/3)(2^3) + 2(2^2) + 5(2) + C

2/3 = 12 + 8/3 + 8 + 10 + C

2/3 = 32 2/3 + C

C = -32

f ' (x) = (3/4)x^4 + (1/3)x^3 + 2x^2 + 5x - 32

To find f(x), integrate f ' (x)

Therefore f(x) = (3/20)x^5 + (1/12)x^4 + (2/3)x^3 + (5/2)x^2 - 32x + C

Now substitute 10 for f(x) and 3 for x

10 = (3/20)(3^5) + (1/12)(3^4) + (2/3)(3^3) + (5/2)(3^2) - 32(3) + C

10 = 729/20 + 81/12 + 18 + 45/2 - 96 + C

10 = -12.3 + C

22 3/10 = C

f(x) = (3/20)x^5 + (1/12)x^4 + (2/3)x^3 + (5/2)x^2 - 32x + 22 3/10.

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