If U_n (n=1,2) denotes the n^{text{th}} deriv | vedclass

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(B) Given $U(x) = \frac{Lx+M}{x^2-2Bx+C}$.Rearranging, we get $U(x)(x^2-2Bx+C) = Lx+M$.Differentiating both sides with respect to $x$ using the produc

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$P=x^2-2Bx+C, Q=4(x-B), R=2$

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(B) Given $U(x) = \frac{Lx+M}{x^2-2Bx+C}$.Rearranging, we get $U(x)(x^2-2Bx+C) = Lx+M$.Differentiating both sides with respect to $x$ using the product rule:$U_1(x^2-2Bx+C) + U(2x-2B) = L$.Differentiating again with respect to $x$:$U_2(x^2-2Bx+C) + U_1(2x-2B) + U_1(2x-2B) + U(2) = 0$.Simplifying the expression:$U_2(x^2-2Bx+C) + U_1(4x-4B) + 2U = 0$.Comparing this with $PU_2 + QU_1 + RU = 0$, we get:$P = x^2-2Bx+C$, $Q = 4(x-B)$, and $R = 2$.

If $U_n$ $(n=1,2)$ denotes the $n^{\text{th}}$ derivative of $U(x) = \frac{Lx+M}{x^2-2Bx+C}$ (where $L, M, B, C$ are constants), then the equation $PU_2 + QU_1 + RU = 0$ holds for:

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