Let f:[2,5] rightarrow R be a differentiable | vedclass

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(A) Given the condition $x f^{\prime}(x) = 2 f(x) - 2 x^3$ for $x \in (2, 5)$.Rearranging the terms, we get the linear differential equation: $\frac{d

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$-2 x^3+\frac{78}{7} x^2$

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(A) Given the condition $x f^{\prime}(x) = 2 f(x) - 2 x^3$ for $x \in (2, 5)$.Rearranging the terms, we get the linear differential equation: $\frac{d y}{d x} - \frac{2}{x} y = -2 x^2$, where $y = f(x)$.The integrating factor is $IF = e^{\int -\frac{2}{x} d x} = e^{-2 \ln x} = \frac{1}{x^2}$.Multiplying both sides by the integrating factor: $\frac{d}{d x} \left( \frac{y}{x^2} \right) = -2$.Integrating both sides with respect to $x$: $\frac{y}{x^2} = -2 x + C$, which gives $f(x) = -2 x^3 + C x^2$.Given $\frac{f(5)}{f(2)} = 1$, we have $f(5) = f(2)$.Substituting $x = 5$ and $x = 2$: $-2(125) + C(25) = -2(8) + C(4)$.$-250 + 25 C = -16 + 4 C$.$21 C = 234 \Rightarrow C = \frac{234}{21} = \frac{78}{7}$.Thus, $f(x) = -2 x^3 + \frac{78}{7} x^2$.

Let $f:[2,5] \rightarrow R$ be a differentiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in (2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$

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