Question

Решить дифференциальные ! 1) (1+x)dx+(y-x^2*y)dy=0 2) (1+2y)*xdx+(1+x^2)*dy=0

Answer 1

$1)\; \; (1+x)dx+(y-x^2y)dy=0\\\\(1+x)dx+y(1-x^2)dy=0\\\\\int \frac{1+x}{1-x^2}dx=-\int y\, dy\\\\\ \int \frac{dx}{1-x}=-\int y\, dy\\\\-ln|1-x|=-\frac{y^2}{2} +C_1\\\\y^2=2ln|1-x|+C\; ,\; \; C=2C_1\\\\2)\; \; (1+2y)x\, dx+(1+x^2)dy=0\\\\\int \frac{x\, dx}{1+x^2}=-\int \frac{dy}{1+2y} \\\\ \frac{1}{2} \, \int \frac{2x\, dx}{1+x^2}=-\frac{1}{2} \, \int \frac{2\, dy}{1+2y} \; \; ,\quad \Big [\; 2x\, dx=d(1+x^2)\; ,\; 2\, dy=d(1+2y)\; \Big ] \\\\ \frac{1}{2}\cdot ln|1+x^2|=-\frac{1}{2} \cdot ln|1+2y|-\frac{1}{2}\, lnC$

$ln(1+x^2)=-ln\Big (C(1+2y)\Big )\\\\1+x^2=\frac{1}{C(1+2y)}\\\\1+2y= \frac{1}{C(1+x^2)} \\\\y= \frac{1}{2C(1+x^2)} -\frac{1}{2}$