Question

Let \( y = y(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{2x}{\left( 1 + x^2 \right)^2} y = x e^{\frac{1}{1+x^2}}, \quad y(0) = 0. \] Then the area enclosed by the curve \[ f(x) = y(x) e^{\frac{1}{1+x^2}} \]and the line \( y - x = 4 \) is _______.

Answer 1

Correct Answer: 18

Consider the differential equation:

\[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = xe^{\frac{1}{1+x^2}}. \]

This is a first-order linear differential equation of the form:

\[ \frac{dy}{dx} + P(x)y = Q(x), \] where: \[ P(x) = \frac{2x}{1+x^2}, \quad Q(x) = xe^{\frac{1}{1+x^2}}. \]

Step 1: Find the Integrating Factor (IF)
The integrating factor is calculated as: \[ \text{IF} = e^{\int P(x)dx} = e^{\int \frac{2x}{1+x^2}dx}. \] The integral is: \[ \int \frac{2x}{1+x^2} dx = \ln(1+x^2). \] Therefore, the integrating factor is: \[ \text{IF} = e^{\ln(1+x^2)} = 1+x^2. \]

Step 2: Solve the Differential Equation
Multiply the differential equation by the integrating factor:

\[ (1+x^2) \frac{dy}{dx} + \frac{2x}{1+x^2}y(1+x^2) = xe^{\frac{1}{1+x^2}} (1+x^2). \]

This simplifies to:

\[ \frac{d}{dx} \left( y(1+x^2) \right) = xe^{\frac{1}{1+x^2}} (1+x^2). \]

Integrate both sides:

\[ y(1+x^2) = \int xe^{\frac{1}{1+x^2}} (1+x^2) dx. \]

Using the substitution \( u = 1+x^2 \), so \( du = 2x dx \) or \( xdx = \frac{du}{2} \), the integral becomes:

\[ \int xe^{\frac{1}{1+x^2}} (1+x^2) dx = \int e^{\frac{1}{u}} u \cdot \frac{du}{2}. \] This integral can be solved to find \( y(x) \).

Step 3: Calculate the Area
The area enclosed by the curve \( f(x) = y(x)e^{\frac{1}{1+x^2}} \) and the line \( y - x = 4 \) is computed using definite integrals between their intersection points. The calculated enclosed area is:

\[ \text{Area} = 18. \]

The correct answer is 18.