Let a curve y=y(x) passing through the points (0,1) and (1,0) satisfy the differential equation (x^2+1) dy dx + k x (y^2+1) = 0 , where k is a constant. Then the value of e^ - y'(1) is equal to:

Step-by-step solution

Given differential equation: (x^2+1) dy dx + k x (y^2+1) = 0 Separating the variables, we get: dy y^2+1 + kx x^2+1 dx = 0 Integrating both sides: dy y^2+1 + k x x^2+1 dx = 0 ⁻¹ y + k 2 (x^2+1) = C Since the curve passes through (0,1) , we substitute x=0, y=1 : ⁻¹(1) + k 2 (1) = C C = 4 The curve also passes through (1,0) , so we substitute x=1, y=0 : ⁻¹(0) + k 2 (2) = 4 k 2 2 = 4 k = 2 2 To find y'(1) , we can directly substitute x=1 and y=0 into the original differential equation: (1^2+1) y'(1) + k(1)(0^2+1) = 0 2