Question

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

• A. \(P = x^2 - 2B, Q = 2x, R = 3x\)
• B. \(P = x^2 - 2Bx + C, Q = 4(x - B), R = 2\)
• C. \(P = 2x, Q = 2B, R = 2\)
• D. \(P = x, Q = x, R = 3\)

Hint:

When dealing with derivatives of rational functions, use the quotient rule, and when solving for related constants, match the degree of terms on both sides of the equation.

Answer 1

The Correct Option is B

Step 1: The given function is \( U(x) = \frac{Lx + M}{x^2 - 2Bx + C} \). We are required to find the relations between the terms \( P, Q, R \) for the equation \( PU_2 + QU_1 + RU = 0 \), where \( U_1 \) and \( U_2 \) are the first and second derivatives of \( U(x) \).

Step 2: First, compute the first and second derivatives of \( U(x) \). Use the quotient rule for differentiation:

\[ U_1(x) = \frac{(x^2 - 2Bx + C)(L) - (Lx + M)(2x - 2B)}{(x^2 - 2Bx + C)^2} \]

Then, compute the second derivative \( U_2(x) \).

Step 3: Now, substitute \( U_1(x) \) and \( U_2(x) \) into the equation \( PU_2 + QU_1 + RU = 0 \).

Step 4: After solving for \( P, Q, R \), we find that the correct values are:

\[ P = x^2 - 2Bx + C, \quad Q = 4(x - B), \quad R = 2 \]

Answer 2

Finding Coefficients of a Differential Equation

Given the function $U(x) = \frac{Lx + M}{x^2 - 2Bx + C}$, where $L, M, B, C$ are constants, and $U_n$ denotes the $n^{th}$ derivative of $U(x)$ ($n = 1, 2$). We need to find the values of $P, Q, R$ for which the equation $PU_2 + QU_1 + RU = 0$ holds.

Step 1: Express the given function

We have $U(x) = \frac{Lx + M}{x^2 - 2Bx + C}$. This can be rewritten as:

$U(x)(x^2 - 2Bx + C) = Lx + M$

Step 2: Differentiate with respect to $x$ (First Derivative)

Differentiating both sides with respect to $x$ using the product rule:

$\frac{d}{dx}(U(x)(x^2 - 2Bx + C)) = \frac{d}{dx}(Lx + M)$

$U_1(x)(x^2 - 2Bx + C) + U(x)(2x - 2B) = L$

Step 3: Differentiate with respect to $x$ again (Second Derivative)

Differentiating the equation from Step 2 with respect to $x$ again using the product rule:

$\frac{d}{dx}(U_1(x)(x^2 - 2Bx + C)) + \frac{d}{dx}(U(x)(2x - 2B)) = \frac{d}{dx}(L)$

$U_2(x)(x^2 - 2Bx + C) + U_1(x)(2x - 2B) + U_1(x)(2x - 2B) + U(x)(2) = 0$

Step 4: Simplify the equation

Combine like terms in the equation from Step 3:

$U_2(x)(x^2 - 2Bx + C) + 2U_1(x)(2x - 2B) + 2U(x) = 0$

$U_2(x)(x^2 - 2Bx + C) + 4U_1(x)(x - B) + 2U(x) = 0$

Step 5: Compare with the given form $PU_2 + QU_1 + RU = 0$

By comparing the coefficients of $U_2(x)$, $U_1(x)$, and $U(x)$ in the derived equation with the given form, we can identify $P, Q, R$:

$P = x^2 - 2Bx + C$

$Q = 4(x - B)$

$R = 2$

This matches option (B).

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