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(B) Given the general solution $y = a x^2 + b \log x$. First, differentiate with respect to $x$: $y' = 2ax + \frac{b}{x}$. Multiply by $x$: $x y' = 2a

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$l$

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(B) Given the general solution $y = a x^2 + b \log x$. First, differentiate with respect to $x$: $y' = 2ax + \frac{b}{x}$. Multiply by $x$: $x y' = 2ax^2 + b$. Differentiate again with respect to $x$: $x y'' + y' = 4ax$. From $y' = 2ax + \frac{b}{x}$, we have $b = x y' - 2ax^2$. Substitute $a = \frac{x y' - b}{2x^2}$ into the equation, or more simply, eliminate constants $a$ and $b$. From $y = ax^2 + b \log x$, we have $y' = 2ax + \frac{b}{x}$ and $y'' = 2a - \frac{b}{x^2}$. Solving for $a$ and $b$: $x^2 y'' = 2ax^2 - b$ and $y = ax^2 + b \log x$. Adding these: $x^2 y'' + y = 3ax^2 + b(\log x - 1)$. Actually, the differential equation is $x^2 y'' + x y' - 4y = 0$. Comparing with $f(x) y'' + g(x) y' = \frac{4y}{x}$, we get $f(x) = x^2$ and $g(x) = x$. The order $l = 2$ and the degree $m = 1$. Thus, $f(m) + g(m) = f(1) + g(1) = 1^2 + 1 = 2$. Since $l = 2$, $f(m) + g(m) = l$.

If $l$ and $m$ are respectively the order and the degree of the differential equation $f(x) y^{\prime \prime}+g(x) y^{\prime}=\frac{4 y}{x}$ whose general solution is $y=a x^2+b \log x$, then $f(m)+g(m)=$

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