Question

Решите парочку поизз

Answer 1

$1)\; \; \sqrt{2+x^2}\, dy+\sqrt{3+y^2}\, dx=0\\\\\int \frac{dy}{\sqrt{3+y^2}}=-\int \frac{dx}{\sqrt{2+x^2}}\\\\ln|y+\sqrt{3+y^2}|=-ln|x+\sqrt{2+x^2}|+lnC\\\\(y+\sqrt{3+y^2})\cdot (x+\sqrt{2+x^2})=C$

$2)\; \; e^{-x}\cdot y'+cos^2y=0\\\\\frac{dy}{dx}=-\frac{cos^2y}{e^{-x}}\\\\\int \frac{dy}{cos^2y}=-\int e^{x}\, dx\\\\tgy=-e^{x}+C$

$3)\; \; 2y^4-3xy'=0\\\\y'=\frac{2y^4}{3x}\\\\\int \frac{dy}{y^4}=\frac{2}{3}\int \frac{dx}{x}\\\\\frac{y^{-3}}{-3}=\frac{2}{3}\cdot ln|x|+C\\\\-\frac{1}{3y^3}=\frac{2}{3}\cdot ln|x|+C\\\\3y^3=-\frac{1}{ln\sqrt[3]{x^2}+C}\\\\y^3=-\frac{1}{3\cdot (ln\sqrt[3]{x^2}+C)}\\\\y=-\frac{1}{\sqrt[3]{3\cdot (ln\sqrt[3]{x^2}+C)}}$