Question

Решить дифференциальное уравнение х2(dу/dx) + (x)1/3 tgy = 0

Answer 1

$x^2y'+dfrac{x}{3}{rm tg}, y=0\ \x=0;~~ xy'+dfrac{{rm tg}, y}{3}=0\ \ xy'=-dfrac{{rm tg}, y}{3}~~~Rightarrow~~~ dfrac{dy}{{rm tg}, y}=-dfrac{dx}{3x}~~~Rightarrow~~ displaystyle int dfrac{cos ydy}{sin y}=-int dfrac{dx}{3x}\ \ \ int dfrac{d(sin y)}{sin y}=-int dfrac{dx}{3x}~~~Rightarrow~~~ ln|sin y|=-dfrac{1}{3}ln|x|+ln C\ \ ln |sin y|=ln left|dfrac{C}{sqrt[3]{x}}right|\ \ sin y=dfrac{C}{sqrt[3]{x}}~~~Rightarrow~~~ boxed{y=arcsindfrac{C}{sqrt[3]{x}}}$