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(A) Given equation: $x^2 f(x) - x = 4 \int_0^x t f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $2x f(x) + x^2 f'(x)

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$160$

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(A) Given equation: $x^2 f(x) - x = 4 \int_0^x t f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $2x f(x) + x^2 f'(x) - 1 = 4x f(x)$. Rearranging the terms: $x^2 f'(x) - 2x f(x) = 1$. Dividing by $x^2$ (assuming $x 
eq 0$): $f'(x) - \frac{2}{x} f(x) = \frac{1}{x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = \frac{1}{x^2}$. Integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2} = \frac{1}{x^2}$. The solution is $f(x) \cdot IF = \int Q(x) \cdot IF dx + C$. $f(x) \cdot \frac{1}{x^2} = \int \frac{1}{x^2} \cdot \frac{1}{x^2} dx = \int x^{-4} dx = \frac{x^{-3}}{-3} + C = -\frac{1}{3x^3} + C$. Multiplying by $x^2$: $f(x) = -\frac{1}{3x} + Cx^2$. Using $f(1) = \frac{2}{3}$: $\frac{2}{3} = -\frac{1}{3} + C(1)^2 \Rightarrow C = 1$. Thus,$f(x) = x^2 - \frac{1}{3x}$. Calculating $18 f(3)$: $f(3) = (3)^2 - \frac{1}{3(3)} = 9 - \frac{1}{9} = \frac{81-1}{9} = \frac{80}{9}$. $18 f(3) = 18 	imes \frac{80}{9} = 2 	imes 80 = 160$.

Let $f$ be a differentiable function such that $x^2 f(x) - x = 4 \int_0^x t f(t) dt$ and $f(1) = \frac{2}{3}$. Then $18 f(3)$ is equal to $......$.

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