Question

Решить уравнение.....

Answer 1

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