Question #128191

(y-zx)p+(x+yz)q=x2+y2

1
Expert's answer
2020-08-03T18:48:07-0400

dxyxz=dyx+yz=dzx2+y2=tdx=ytxzt,dy=xt+yzt,dz=(x2+y2)tydx+xdy=y2txyzt+x2t+xyzt=(x2+y2)tydx+xdyx2+y2=dzx2+y2=tdz=ydx+xdyzx=y,zy=xz(x,y)=xy+c;check:(yxz)p+(x+yz)q=(yx(xy+c))zx+(x+y(xy+c))zy=y2xy(xy+c)+x2+xy(xy+c)=x2+y2;\frac{dx}{y-xz}=\frac{dy}{x+yz}=\frac{dz}{x^2+y^2}=t\\ dx=yt-xzt,\quad dy=xt+yzt,\quad dz=(x^2+y^2)t\\ ydx+xdy=y^2t-xyzt+x^2t+xyzt=(x^2+y^2)t\Longrightarrow\\ \frac{ydx+xdy}{x^2+y^2}=\frac{dz}{x^2+y^2}=t\Longrightarrow\\ dz=ydx+xdy\Longrightarrow z'_x=y,\quad z'_y=x \Longrightarrow z(x,y)=xy+c;\\ check:\\ (y-xz)p+(x+yz)q=(y-x\cdot (xy+c))\cdot z'_x + (x+y\cdot (xy+c))\cdot z'_y=\\ y^2-xy(xy+c)+x^2+xy(xy+c)=x^2+y^2;


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