Question

Y'=xy^2 (equação diferencial separaveis) y (2)=1

Answer 1

$y'=xy^2\\\\ \dfrac{y'}{y^2}=x\\\\ \displaystyle\int\dfrac{dy}{y^2}=\int x\,dx\\\\ \dfrac{y^{-2+1}}{-2+1}=\dfrac{x^2}{2}+C\\\\ -\dfrac{1}{y}=\dfrac{x^2}{2}+C\\\\ \dfrac{1}{y}=-C-\dfrac{x^2}{2}$

Usando a condição de contorno y(2)=1:

$\dfrac{1}{y(x)}=-C-\dfrac{x^2}{2}\\\\ \dfrac{1}{y(2)}=-C-\dfrac{2^2}{2}\\\\ \dfrac{1}{1}=-C-\dfrac{4}{2}\\\\ 1=-C-2\\\\ C=-3$

Então:

$\dfrac{1}{y}=3-\dfrac{x^2}{2}\\\\ </span><span>\dfrac{1}{y}=\dfrac{6-x^2}{2} \\\\ \boxed{y=\dfrac{2}{6-x^2}}$