To solve this problem, we need to find derivatives and analyze the given function f(x) = \int_{0}^{x} t(\sin x - \sin t) \, dt. Let's break down the steps:
First, differentiate f(x) with respect to x. Using the Leibniz rule for differentiation under the integral, we have:
f'(x) = x(\sin x - \sin x) + \int_{0}^{x} t \cos x \, dt = \int_{0}^{x} t \cos x \, dt.
Since \sin x - \sin x = 0, this simplifies to:
f'(x) = \cos x \int_{0}^{x} t \, dt. When integrating, we get:
f'(x) = \cos x \left[ \frac{t^2}{2} \right]_{0}^{x} = \cos x \cdot \frac{x^2}{2} = \frac{x^2}{2} \cos x.
Now, differentiate f'(x) to find f''(x):
f''(x) = \frac{d}{dx} \left(\frac{x^2}{2} \cos x\right) = \frac{x^2}{2} (-\sin x) + x \cos x.
Simplifying, this becomes:
f''(x) = -\frac{x^2}{2} \sin x + x \cos x.
Next, differentiate f''(x) to find f'''(x):
f'''(x) = \frac{d}{dx} [-\frac{x^2}{2} \sin x + x \cos x].
Applying the product rule, we get:
f'''(x) = -\frac{d}{dx} \left(\frac{x^2}{2} \sin x\right) + \frac{d}{dx} (x \cos x) = [-x \sin x - \frac{x^2}{2} \cos x] + [\cos x - x \sin x].
Simplifying, we have:
f'''(x) = -x \sin x - \frac{x^2}{2} \cos x + \cos x - x \sin x = -2x \sin x + \cos x - \frac{x^2}{2} \cos x.
After obtaining f'(x) and f'''(x), we can examine which differential equation corresponds to the options given. We find that the equation:
f'''(x) + f'(x) = \cos x - 2x \sin x
fits the calculations, considering:
f'''(x) = \cos x - 2x \sin x
f'(x) = \frac{x^2}{2} \cos x
Thus, adding these derivatives leads to:
f'''(x) + f'(x) = (-2x \sin x + \cos x - \frac{x^2}{2} \cos x) + \frac{x^2}{2} \cos x = \cos x - 2x \sin x
Therefore, the correct answer is:
Option 3: f'''(x) + f'(x) = \cos x - 2x \sin x
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