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Recovering f(x) from f''(x) via successive antidifferentiation

Suppose a function f(x)f(x) satisfies f′′(x)=12x2−4f''(x) = 12x^2 - 4, with initial conditions f′(0)=3f'(0) = 3 and f(0)=−2f(0) = -2.

Find f(x)f(x) by integrating f′′(x)f''(x) twice, using the given initial conditions f′(0)=3f'(0) = 3 and f(0)=−2f(0) = -2 to determine the constants of integration. State the resulting formula for f(x)f(x).

Hint

First find the general antiderivative of f′′(x)f''(x) to get f′(x)+C1f'(x) + C_1, use f′(0)=3f'(0)=3 to pin down C1C_1. Then integrate f′(x)f'(x) to get f(x)+C2f(x) + C_2, and use f(0)=−2f(0)=-2 to pin down C2C_2. Remember every indefinite integral needs a +C+C before you use the condition to solve for it.

Step-by-step solution

Integrate f′′(x)=12x2−4f''(x) = 12x^2 - 4 term by term: ∫12x2 dx=4x3\int 12x^2\,dx = 4x^3 and ∫−4 dx=−4x\int -4\,dx = -4x, so f′(x)=4x3−4x+C1f'(x) = 4x^3 - 4x + C_1. Using f′(0)=3f'(0) = 3: 4(0)3−4(0)+C1=34(0)^3 - 4(0) + C_1 = 3, so C1=3C_1 = 3. Thus f′(x)=4x3−4x+3f'(x) = 4x^3 - 4x + 3.

Now integrate f′(x)=4x3−4x+3f'(x) = 4x^3 - 4x + 3: ∫4x3 dx=x4\int 4x^3\,dx = x^4, ∫−4x dx=−2x2\int -4x\,dx = -2x^2, and ∫3 dx=3x\int 3\,dx = 3x, giving f(x)=x4−2x2+3x+C2f(x) = x^4 - 2x^2 + 3x + C_2. Using f(0)=−2f(0) = -2: 0−0+0+C2=−20 - 0 + 0 + C_2 = -2, so C2=−2C_2 = -2. Therefore f(x)=x4−2x2+3x−2f(x) = x^4 - 2x^2 + 3x - 2.

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