If x phi(x) = int_{5}^{x} (3t^{2} - 2 phi'(t) | vedclass

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(A) Given the equation: $x \phi(x) = \int_{5}^{x} (3t^{2} - 2 \phi'(t)) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $\

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$4$

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(A) Given the equation: $x \phi(x) = \int_{5}^{x} (3t^{2} - 2 \phi'(t)) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $\frac{d}{dx} [x \phi(x)] = 3x^{2} - 2 \phi'(x)$. Applying the product rule on the left side: $\phi(x) + x \phi'(x) = 3x^{2} - 2 \phi'(x)$. Rearranging the terms to group $\phi'(x)$: $(x + 2) \phi'(x) = 3x^{2} - \phi(x)$. This is a linear differential equation: $\phi'(x) + \frac{1}{x+2} \phi(x) = \frac{3x^{2}}{x+2}$. The integrating factor is $I.F. = e^{\int \frac{1}{x+2} dx} = e^{\ln(x+2)} = x+2$. Multiplying by $I.F.$: $(x+2) \phi'(x) + \phi(x) = 3x^{2}$. Integrating both sides: $(x+2) \phi(x) = \int 3x^{2} dx = x^{3} + C$. Given $\phi(0) = 4$: $(0+2) \phi(0) = 0^{3} + C \Rightarrow 2(4) = C \Rightarrow C = 8$. So,$(x+2) \phi(x) = x^{3} + 8$. $\phi(x) = \frac{x^{3} + 8}{x+2} = \frac{(x+2)(x^{2} - 2x + 4)}{x+2} = x^{2} - 2x + 4$. For $x = 2$: $\phi(2) = 2^{2} - 2(2) + 4 = 4 - 4 + 4 = 4$.

If $x \phi(x) = \int_{5}^{x} (3t^{2} - 2 \phi'(t)) dt$,$x > -2$,and $\phi(0) = 4$,then $\phi(2)$ is .... .

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