Let y=y(x) be the solution of the differential equation
(d y)/(d x)+(2 x)/(1+x²)² y=x e^(1/(1+x²)) ; y(0)=0.
Then the area enclosed by the curve f(x)=y(x) e^(-1/(1+x²)) and the line y-x=4 is ________.

Solution

Maths sir · 5 steps

The integrating factor is exactly the e^(-1/(1+x²)) in f, so f turns out to be a plain parabola.

1. Integrating factor. Here P(x) = 2x/(1+x²)², and ∫ P dx = -1/(1+x²). So the integrating factor is e^(-1/(1+x²)), the same exponential that appears in f(x).

   d/dx(y e^(-1/(1+x²))) = x e^(1/(1+x²)) · e^(-1/(1+x²)) = x

2. Find f. Integrating gives y e^(-1/(1+x²)) = x²/2 + C. At x = 0, y = 0, so C = 0. This means f(x) = x²/2.
3. Meeting points. The line is y = x + 4. Setting x²/2 = x + 4 gives x² - 2x - 8 = 0, so x = -2 and x = 4. Between them the line lies above the parabola.
4. Area. ∫₋₂⁴(x + 4 - x²/2)dx = [x²/2 + 4x - x³/6]₋₂⁴ = 40/3 - (-14/3) = 54/3.
5. So. The enclosed area is 18 square units.

The common slip: 36. Using the shortcut (β - α)³/6 without multiplying by the leading coefficient 1/2 gives 216/6 = 36, not 18.

The idea to keep. When a question builds a new function from y times some exponential, check whether that exponential is the integrating factor. If it is, the new function is just the integral of the right side.

Answer from the NTA final answer key. Solution by PrepWiser.