Question

Найти общий интеграл дифференциального уравнения 1) (x-y*cos(y/x))*dx+x*cos(y/x)*dy=0 2) xy+y^2=(2*x^2+xy)*y'

Answer 1

$(x-y*cos\frac{y}{x})+x*cos\frac{y}{x}*y'=0\\(\lambda x-\lambda y*cos\frac{\lambda y}{\lambda x})+\lambda x*cos\frac{\lambda y}{\lambda x}*y'=0\\(x-y*cos\frac{y}{x})+x*cos\frac{y}{x}*y'=0\\y=tx;y'=t'x+t\\x-xtcost+xcost(t'x+t)=0\\x+x^2costt'=0\\1+xcost\frac{dt}{dx}=0|*\frac{dx}{x}\\costdt=-\frac{dx}{x}\\\int costdt=-\int\frac{dx}{x}\\sint=-ln|x|+C\\sin\frac{y}{x}+ln|x|=C\\$

$(sin\frac{y}{x}+ln|x|)'=C'\\cos\frac{y}{x}*\frac{y'-y}{x^2}+\frac{1}{x}=0|*x^2\\cos\frac{y}{x}(y'-y)+x=0\\x+y'cos\frac{y}{x}-y*cos\frac{y}{x}=0\\(x-ycos\frac{y}{x})dx+ycos\frac{y}{x}dy=0$
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$xy+y^2=(2*x^2+xy)*y'\\\lambda^2xy+\lambda^2y^2=(2*\lambda^2x^2+\lambda^2xy)*y'\\\lambda^2(xy+y^2)=\lambda^2(2*x^2+xy)*y\\xy+y^2=(2*x^2+xy)*y'\\y=tx;y'=t'x+t\\x^2t+x^2t^2=(2x^2+x^2t)(t'x+t)|:x^2\\t+t^2=(2+t)(t'x+t)\\t+t^2=(2+t)t'x+2t+t^2\\-t=(2+t)\frac{dt}{dx}x|*\frac{dx}{xt}\\-\frac{dx}{x}=\frac{(2+t)dt}{t}\\$
При делении теряем решение:
$t=0\\\frac{y}{x}=0\\y=0;y'=0\\x*0+0=(2*x^2+x*0)*0\\0=0$
Запоминаем, едем дальше
$-\int\frac{dx}{x}=\int(2\frac{1}{t}+1)dt\\-ln|x|=2ln|t|+t+C\\ln|x|+ln|t^2|+t=C\\ln|x|+ln|\frac{y^2}{x^2}+\frac{y}{x}=C\\ln|\frac{y^2}{x}|+\frac{y}{x}=C;y=0$

$(ln|\frac{y^2}{x}|+\frac{y}{x})'=C'\\\frac{x}{y^2}*\frac{2yy'x-y^2}{x^2}+\frac{y'x-y}{x^2}=0\\\frac{2y'x-y}{xy}+\frac{y'x-y}{x^2}=0|*x^2y\\2y'x^2-xy+yy'x-y^2=0\\xy+y^2=(2x^2+xy)y'$