Question

$\frac{dy}{\sqrt{x}}=\frac{3dx}{\sqrt{y} }$
$x^{2} dy-(2xy+3y)dx=0$

Answer 1

Ответ: во вложении Пошаговое объяснение:

Answer 2

$1)\; \; \frac{dy}{\sqrt{x}}=\frac{3\, dx}{\sqrt{y}}\; \; \; \Rightarrow \; \; \; \int \sqrt{y}\, dy=3\int \sqrt{x}\, dx\\\\\frac{y^{3/2}}{3/2}=3\cdot \frac{x^{3/2}}{3/2}+C\\\\\frac{2\sqrt{y^3}}{3}=2\, \sqrt{x^3}+C$

$2)\; \; x^2\, dy-(2xy+3y)\, dx=0\\\\x^2\, dy=y\cdot (2x+3)\, dx\\\\\int \frac{dy}{y}=\int \frac{(2x+3)dx}{x^2}\\\\\int \frac{dy}{y}=2\int \frac{dx}{x}+3\int \frac{dx}{x^2}\\\\ln|y|=2ln|x|-\frac{3}{x}+C$