Дифференциальные уравнения: 1)y^2dx+x^2dy=xydy, 2)(x-y)dx+xdy=0, 3)(2x+y+1)dx+(x+2y-1)dy=0.

Ответы

kisel2014 05.10.2020 05:12

$1)\; \; y^2\, dx+x^2\, dy=xy\, dy\\\\y^2\, dx=(xy-x^2)dy\\\\\frac{dy}{dx}=\frac{y^2}{xy-x^2}\\\\u=\frac{y}{x}\; ,\; y=u\cdot x\; ,\; \; y'=u'x+u\\\\\frac{dy}{dx}=y'\; ,\; \; u'x+u=\frac{u^2x^2}{ux^2-x^2}\\\\u'x+u=\frac{u^2}{u-1}\\\\u'x=\frac{u^2}{u-1}-u\; ;\; \; \; u'x=\frac{u^2-u^2+u}{u-1}\; ;\; \; \; u'x=\frac{u}{u-1}\; ;\\\\u'=\frac{du}{dx}\; ,\; \; \frac{du}{dx}=\frac{u}{u-1}\; ;\; \; \frac{(u-1)du}{u}=dx\\\\\int (1-\frac{1}{u})du=\int dx\\\\u-ln|u|=x+C\\\\\frac{y}{x}-ln\left |\frac{y}{x}\right |=x+C$

$2)\; \; \; (x-y)dx+x\, dy=0\\\\(x-y)dx=-x\, dy\\\\\frac{dy}{dx}=\frac{x-y}{-x}\; ;\; \; \frac{dy}{dx}= \frac{y-x}{x} \; ;\; \; \frac{dy}{dx}=\frac{y}{x}-1\; ;\\\\u=\frac{y}{x}\; ,y=ux\; ,\; y'=u'x+u\\\\u'x+u=u-1\\\\u'x=-1\; ;\; \; \frac{du}{dx}\cdot x=-1\; ;\; \; \int du=-\int \frac{dx}{x}\; ;\\\\u=-ln|x|+C\\\\\frac{y}{x}=-ln|x|+C$

$3)\; \; (2x+y+1)dx+(x+2y+1)dy=0\\\\P=2x+y+1\; ,\; P'_{y}=1\\\\Q=x+2y-1\; ,\; \; Q'_{x}=1\\\\P'_{y}=Q'_{x}\; \; \to \; \; F'_{x}dx+F'_{y}dy=0\\\\F'_{x}=2x+y+1\; ;\; \; F'_{y}=x+2y-1\\\\F=\int (2x+y+1)dx=x^2+yx+x+\varphi(y)\\\\F'_{y}=x+\varphi '_{y}(y)=x+2y-1\\\\\varphi '_{y}(y)=2y-1\; \; \to \; \; \varphi (y)=\int (2y-1)dy=y^2-y+C\\\\F=x^2+xy+x+(y^2-y+C)\; \; \to \\\\x^2+y^2+xy+x-y+C_1=0\; \; obshij\; integral$