Question

Resolver as equações diferenciais:
$y^{2}dx - (1 - x)dy = 0$
__ separável​

Answer 1

$y^2dx-(1-x)dy=0\ \therefore\ y^2dx=(1-x)dy\ \therefore\\\\\ \dfrac{dx}{1-x}=\dfrac{dy}{y^2}\ \therefore\ \int{\dfrac{dx}{1-x}}=\int{\dfrac{dy}{y^2}}$

$u=1-x\ \therefore\ \dfrac{du}{dx}=-1\ \therefore\ dx=-du\\\\ \int{\dfrac{dx}{1-x}}=\int{\dfrac{dy}{y^2}}\ \therefore\ \int{-\dfrac{du}{u}}=\int{y^{-2}dy}\ \therefore$

$-(\ln{u}+C_1)=\dfrac{y^{-1}}{(-1)}-C_2\ \therefore\ \ln{(1-x)}+C_1=\dfrac{1}{y}+C_2\ \therefore$

$\dfrac{1}{y}=\ln{(1-x)}+C\ \therefore\ \boxed{y=\dfrac{1}{\ln{(1-x)}+C}}$

Não esqueça de classificar como a melhor resposta!