We have an answer to "Solve (y+1)dx+(x+1)dy=0"

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Answered question

2022-04-20

Solve

(y + 1) d x + (x + 1) d y = 0

Answer & Explanation

Elliot Roy

Beginner2022-04-21Added 14 answers

Step 1
Simpler:

(y + 1) d x + (x + 1) d y = 0

y d x + x d y = - (d x + d y)

d (x y) = d (- (x + y))

x y = - (x + y) + c

x y + x + y = c

,
which is equivalent to your given answer.

Deandre Barron

Beginner2022-04-22Added 17 answers

Step 1
$(y + 1) d x + (x + 1) d y = 0$
$M (x, y) = y + 1$
$N (x, y) = x + 1$
Then, $\frac{\partial M}{\partial y} = 1 = \frac{\partial N}{\partial x}$
Hence, the differential equation is exact.
Choose, $u (x, y)$ be such that
$u_{x} = M$ an d $u_{y} = N$
Then, $u (x, y) = \int {M {d x}} = x (y + 1) + h (y)$
$u_{y} (x, y) = x + h^{'} (y)$
$x + h^{'} (y) = x + 1$
$h (y) = y$
Hence, $u (x, y) = x (y + 1) + y$
Solution: $u (x, y) = C$
$\Rightarrow x (y + 1) + y = C$
$\Rightarrow x (y + 1) + y + 1 = C + 1$
$(y + 1) (x + 1) = [c = C + 1]$
Hence, $(y + 1) = \frac{c}{(x + 1)}$
Note:
$\frac{d}{d x} {u (x, y (x))}$
$= u_{x} + u_{y} \frac{d y}{d x}$
$= M + N \frac{d y}{d x} = 0$
Hence, $u (x, y) = C$

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