Question Number 104357 by bemath last updated on 21/Jul/20

(x+y+1) (dy/dx) = 1

Answered by john santu last updated on 21/Jul/20

let z = x+y+1 (dz/dx) = 1+ (dy/dx) ⇒(dy/dx) = (dz/dx)−1 (→) z.((dz/dx)−1) = 1 (→) (dz/dx) = (1/z)+1 (→) (dz/dx) = ((1+z)/z) ; ((z dz)/(1+z)) = dx (→) ∫ (((1+z−1)dz)/(1+z)) = x +C (→) ∫ dz −∫ (dz/(z+1)) = x +C z− ln ∣z+1∣ = x +C ∴ x+y+1−ln ∣x+y+2∣ = x+C y − ln ∣x+y+2∣ = K (JS ⊛)

Answered by OlafThorendsen last updated on 21/Jul/20

x+y+1 = (dx/dy) (dx/dy)−x = y+1 x_P = −y+b ⇒ −1+y−b = y+1 b = −2 x_P = −y−2 (dx_H /dy)−x_H = 0 (dx_H /x_H ) = dy ln∣x_H ∣ = y+C x_H = Ke^y Finally x = x_P +x_H = Ke^y −y−2 In this kind of probem x = x(y) is better than y = y(x)