x-y-1-dy-dx-1-

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

(x + y + 1) \frac{d y}{d x} = 1

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

l e t z = x + y + 1

\frac{d z}{d x} = 1 + \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{d z}{d x} - 1

(\to) z . (\frac{d z}{d x} - 1) = 1

(\to) \frac{d z}{d x} = \frac{1}{z} + 1

(\to) \frac{d z}{d x} = \frac{1 + z}{z}; \frac{z d z}{1 + z} = d x

(\to) \int \frac{(1 + z - 1) d z}{1 + z} = x + C

(\to) \int d z - \int \frac{d z}{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

(J S ⊛)

Answered by OlafThorendsen last updated on 21/Jul/20

x + y + 1 = \frac{d x}{d y}

\frac{d x}{d y} - x = y + 1

x_{P} = - y + b

\Rightarrow - 1 + y - b = y + 1

b = - 2

x_{P} = - y - 2

\frac{{d x}_{H}}{d y} - x_{H} = 0

\frac{{d x}_{H}}{x_{H}} = d y

\ln ∣ x_{H} ∣ = y + C

x_{H} = K e^{y}

Finally x = x_{P} + x_{H} = K e^{y} - y - 2

In this kind of probem

x = x (y) is better than y = y (x)