Find f. f''(θ) = sin(θ) + cos(θ), f(0) = 5, f '(0) = 1

Solution:

It is given that

f''(θ) = sin(θ) + cos(θ)

f’(θ) = ∫f''(θ) dθ = ∫sin(θ) + cos(θ) dθ

f’(θ) = -cos θ + sin θ + C1

It is given that f’(0) = 1

1 = -cos 0 + sin 0 + C1

C1 = 2

f(θ) = ∫f'(θ) dθ = ∫-cos(θ) + sin(θ) + 2 dθ

f(θ) = -sin θ - cos θ + 2θ + C2

It is given that f(0) = 5

5 = -sin (0) - cos (0) + 2(0) + C2

C2 = 6

So we get

f(θ) = -sin θ - cos θ + 2θ + 6

Therefore, f (θ) = -sin θ - cos θ + 2θ + 6.

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Find f. f''(θ) = sin(θ) + cos(θ), f(0) = 5, f '(0) = 1

Summary:

If f''(θ) = sin(θ) + cos(θ), f(0) = 5, f '(0) = 1, f (θ) = -sin θ - cos θ + 2θ + 6.