Question

If \(y=A(x)\cdot e^{-x^2\) is a solution of the differential equation} \[ \frac{dy}{dx}+2xy=2e^{-x^2} \] with \(y(0)=-3\), then the value of \(A(1)\) is ____.

• A. \(-1\)
• B. \(-2\)
• C. \(-3\)
• D. \(2\)

Hint:

For equations of the form \[ \frac{dy}{dx}+P(x)y=Q(x), \] use \[ \boxed{ \text{I.F.}=e^{\int P(x)\,dx}. } \]

Answer 1

The Correct Option is A

Concept: The given differential equation is a first-order linear differential equation. Using the integrating factor, \[ \boxed{ \text{I.F.}=e^{\int 2x\,dx}=e^{x^2} } \] Multiplying the equation by the integrating factor converts the left-hand side into an exact derivative.

Step 1: Multiply both sides by the integrating factor. \[ e^{x^2}\frac{dy}{dx} +2xe^{x^2}y = 2. \] Hence, \[ \frac{d}{dx}\left(ye^{x^2}\right)=2. \] Integrating, \[ ye^{x^2}=2x+C. \] Therefore, \[ y=(2x+C)e^{-x^2}. \] Thus, \[ A(x)=2x+C. \]

Step 2: Use the initial condition. Given, \[ y(0)=-3. \] Hence, \[ C=-3. \] Therefore, \[ A(x)=2x-3. \] Thus, \[ A(1)=2(1)-3=-1. \] Hence, \[ \boxed{A(1)=-1.} \] Therefore, the correct option is \[ \boxed{(A)\;-1.} \]