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

Question

Алгебра

Tianmazh

7 год назад

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

Ираклиевич

Jul 4, 2019

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

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

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

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