Question

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

Answer 1

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

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

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

Answer 2

Спасибо)