Let y = y(x) be the solution of the differential equation dy dx = y^2 + 2y + 2 x^2 + 4x + 5 such that the solution curve passes through the origin. Then _ x y(x) is equal to:

Step-by-step solution

The given differential equation is dy dx = y^2 + 2y + 2 x^2 + 4x + 5 . Rewriting the quadratics by completing the square: dy (y + 1)^2 + 1 = dx (x + 2)^2 + 1 Integrating both sides: dy (y + 1)^2 + 1 = dx (x + 2)^2 + 1 ⁻¹(y + 1) = ⁻¹(x + 2) + C The curve passes through the origin, so y(0) = 0 . Substituting x = 0 and y = 0 : ⁻¹(1) = ⁻¹(2) + C C = 4 - ⁻¹(2) Substituting C back into the equation: ⁻¹(y + 1) = ⁻¹(x + 2) + 4 - ⁻¹(2) Taking the limit as x : _ x ⁻¹(x + 2) = 2 Let L = _ x y(x) . Then: ⁻¹(L + 1) = 2 + 4 - ⁻¹