Question

Решить уравнение.....

Answer 1

$y'=frac{sqrt{x} }{sqrt{y} }\frac{dy}{dx} =frac{sqrt{x} }{sqrt{y} }\sqrt{y}dy=sqrt{x}dx\int{sqrt{y}dy} =int{sqrt{x}dx}\frac{2}{3}y^{3/2}=frac{2}{3}x^{3/2}+C\y^{3/2}=x^{3/2}+C=>y=(x^{3/2}+C)^{2/3}$