Question

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

$(sinfrac{y}{x}+ln|x|)'=C'\cosfrac{y}{x}*frac{y'-y}{x^2}+frac{1}{x}=0|*x^2\cosfrac{y}{x}(y'-y)+x=0\x+y'cosfrac{y}{x}-y*cosfrac{y}{x}=0\(x-ycosfrac{y}{x})dx+ycosfrac{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
$-intfrac{dx}{x}=int(2frac{1}{t}+1)dt\-ln|x|=2ln|t|+t+C\ln|x|+ln|t^2|+t=C\ln|x|+ln|frac{y^2}{x^2}+frac{y}{x}=C\ln|frac{y^2}{x}|+frac{y}{x}=C;y=0$

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