Let y = y(x) be the solution of the differential equation dy dx = x+y+1 x+y-1 , such that y(1) = 0 . If (x₀, y₀) is a point on the curve such that x₀ + y₀ = e^2 , then the value of (y₀ - x₀) is equal to

Step-by-step solution

Given differential equation is dy dx = x+y+1 x+y-1 . Let v = x + y . Then dv dx = 1 + dy dx dy dx = dv dx - 1 . Substituting this into the differential equation: dv dx - 1 = v+1 v-1 dv dx = v+1 v-1 + 1 = v+1+v-1 v-1 = 2v v-1 Separating the variables: v-1 2v dv = dx Integrating both sides: 1 2 (1 - 1 v ) dv = dx 1 2 (v - |v|) = x + C Substitute back v = x + y : 1 2 (x + y - |x+y|) = x + C x + y - |x+y| = 2x + 2C y - x - |x+y| = C' (where C' = 2C ) Given y(1) = 0 , substitute x = 1 and y = 0 : 0 - 1 - |1+0| = C' C' =