Question

Даю 100 балов за все заденея кроме 5!!!!!

Answer 1

Решение во вложении...

Answer 2

$1)\; \; 6xy=(x^2-5)y'\\\\\frac{dy}{dx}=\frac{6xy}{x^2-5}\; \; ,\; \; \int \frac{dy}{3y}=\int \frac{2x\, dx}{x^2-5}\\\\\frac{1}{3}\cdot ln|y|=ln|x^2-5|+C\; ,\\\\ ln\Big |\frac{x^2-5}{\sqrt[3]{y}}\Big |+C=0\\\\\\2)\; \; y'\cdot cosx-y\, sinx=0\; ,\; \; \y(\pi )=-1\\\\\frac{dy}{dx}=\frac{y\, sinx}{cosx}\; \; ,\; \; \int \frac{dy}{y}=\int \frac{sinx\, dx}{cosx}\; \; ,\\\\ln|y|=-ln|cosx|+ln|C|\; ,\; \; \; \; ln(y\cdot cosx)=lnC\\\\y\cdot cosx=C\; \; \to \; \; y=\frac{C}{cosx}\\\\b)\; \; -1=\frac{C}{cos\pi }\; \; \to C=1\\\\y=\frac{1}{cosx}$

$3)\; \; y^{\frac{3}{5}}\, dy=cosx\, dx\\\\\int y^{\frac{3}{5}}\, dy=\int cosx\, dx\; \; \to \; \; \frac{5\, y^{\frac{8}{5}}}{8}=sinx+C_1\; \; ,\\\\y^{\frac{8}{5}}=\frac{8}{5}\, sinx+C\; \; ,\; \; C=\frac{8}{5}C_1\\\\y=\sqrt[8]{(\frac{8}{5}\, sinx+C)^5}$

$4)\; \; y'-xy^2=3xy\; \; ,\; \; y(0)=2\\\\\frac{dy}{dx}=xy(y+3)\; \; ,\; \; \int \frac{dy}{y(y+3)}=\int x\, dx\; ,\\\\\int \frac{dy}{y(y+3)}=\frac{1}{3}\int \frac{dy}{y}-\frac{1}{3}\int \frac{dy}{y+3}=\frac{1}{3}\cdot ln|y|-\frac{1}{3}\cdot ln|y+3|+C\; ;\\\\\frac{1}{3}\cdot \ln\Big |\frac{y}{y+3}\Big |=\frac{x^2}{2}+C\\\\b)\; \; y(0)=2:\; \; \frac{1}{3}\cdot ln\frac{2}{5}=0+C\; ,\; \; C=\frac{1}{3}\, ln\frac{2}{5}\\\\\frac{1}{3}ln\Big |\frac{y}{y+3}\big |=\frac{x^2}{2}+\frac{1}{3}\, ln\frac{2}{5}$