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)