WBJEE · Maths · Differential Equations
If \(U_n(n=1,2)\) denote the \(n^{\text {th }}\) derivative \((n=1,2)\) of \(U(x)=\frac{L x+M}{x^2-2 B x+C}(L, M, B, C\) are constants \()\), then \(P U_2+Q U_1\) \(+R U=0\), holds for
- A \(\mathrm{P}=\mathrm{x}^2-2 \mathrm{~B}, \mathrm{Q}=2 \mathrm{x}, \mathrm{R}=3 \mathrm{x}\)
- B \(P=x^2-2 B x+C, Q=4((x-B), R=2\)
- C \(P=2 x, Q=2 B, R=2\)
- D \(P=x^2, Q=x, R=3\)
Answer & Solution
Correct Answer
(B) \(P=x^2-2 B x+C, Q=4((x-B), R=2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Hint: } U(x)=\frac{L x+m}{x^2-2 B x+c}, U(x)\left(x^2-2 B x+c\right)=L x+m \\ & U(x)(2 x-2 B)+\left(x^2-2 B x+c\right) U_1(x)=L \\ & U(x)2 \times 2(2 x-2 B) U_1(x)+(2 x-2 B) U_1(x)+\left(x^2-2 B x+c\right) U_2(x)=0 \\ & \Rightarrow U_2\left(x^2-2…
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