Question

If \[ \frac{dy}{dx} - \frac{2x}{x^2+b}y = -2x(x^2+b), \] with \[ y(0)=12,\qquad y(1)=10, \] then the sum of all possible values of \(b\) is:

• A. \(1\)
• B. \(4\)
• C. \(-3\)
• D. \(-1\)

Hint:

For linear differential equations, always compute the integrating factor first. It converts the left-hand side into an exact derivative.

Answer 1

The Correct Option is A

Concept: This is a linear differential equation of the form \[ \frac{dy}{dx}+P(x)y=Q(x). \] We solve it using the integrating factor method.

Step 1: Identify \(P(x)\).
Given \[ \frac{dy}{dx} - \frac{2x}{x^2+b}y = -2x(x^2+b). \] Thus \[ P(x) = -\frac{2x}{x^2+b}. \]

Step 2: Find the integrating factor.
\[ IF = e^{\int P(x)\,dx}. \] \[ = e^{\int -\frac{2x}{x^2+b}\,dx}. \] \[ = e^{-\ln(x^2+b)}. \] \[ = \frac1{x^2+b}. \]

Step 3: Multiply throughout by the integrating factor.
\[ \frac1{x^2+b}\frac{dy}{dx} - \frac{2x}{(x^2+b)^2}y = -2x. \] The left side becomes \[ \frac{d}{dx} \left( \frac{y}{x^2+b} \right). \] Hence \[ \frac{d}{dx} \left( \frac{y}{x^2+b} \right) = -2x. \]

Step 4: Integrate.
\[ \frac{y}{x^2+b} = -x^2+C. \] Therefore \[ y=(x^2+b)(C-x^2). \]

Step 5: Use \(y(0)=12\).
\[ 12=bC. \] Thus \[ C=\frac{12}{b}. \]

Step 6: Use \(y(1)=10\).
\[ 10=(1+b)\left(\frac{12}{b}-1\right). \] \[ 10b=(1+b)(12-b). \] \[ 10b=12+11b-b^2. \] \[ b^2-b+12=0. \] \[ (b-4)(b+3)=0. \] Hence \[ b=4 \quad\text{or}\quad b=-3. \]

Step 7: Find the required sum.
\[ 4+(-3)=1. \] Hence \[ \boxed{1}. \]