Question

Resolver a equação de Bernoulli:
dy/dx = sen(x + y)

Answer 1

z = x+y
dz/dx = 1 + dy/dx ---> dy/dx = dz/dx-1

fazendo a substituição
$\frac{dz}{dx}-1 = sen(z)\\\\ \frac{dz}{sen(z)+1} = dx\\\\ \boxed{\boxed{\int \frac{dz}{sen(z)+1} = x}}$

resolvendo a integral usando a substituição universal

$\bmatrix t=tg( \frac{z}{2})\\\\ 2\;arctg(t)=z\\\\ \frac{2dt}{t^2+1}=dz\\\\ sen(z)= \frac{2t}{t^2+1} \end$

temos

$\int \frac{1}{ \frac{2t}{t^2+1} +1}* \frac{2dt}{t^2+1} = \int \frac{1}{ \frac{2t+t^2+1}{t^2+1}}* \frac{2dt}{t^2+1} = \int \frac{2}{t^2+2t+1} dt = 2\int \frac{1}{(t+1)^2} dt = \frac{-2}{t+1}$

então
$\frac{-2}{t+1}+C= x \\\\ \frac{-2}{tg( \frac{z}{2} )+1}+C= x \\\\ \frac{-2}{tg( \frac{x+y}{2} )+1}+C= x$