Question #205784

Solve;  (𝑧 2 − 2𝑦𝑧 − 𝑦 2 )𝑝 + (𝑥𝑦 + 𝑧𝑥)𝑞 = 𝑥𝑦 − 𝑧x


1
Expert's answer
2021-06-14T11:14:47-0400

(z2−2yz−y2)zx+(xy+zx)zy=xy−zxBy Charpit-Lagrange method :dxz2−2yz−y2=dyxy+zx=dzxy−zxFirst characteristic equation,dyxy+zx=dzxy−zxdzdy=y−zy+zz2+2yz−y2=c1Second characteristic equation,xdx+ydy+zdzx(z2−2yz−y2)+y(xy+zx)+z(xy−zx)=0xdx+ydy+zdz0=0  ⟹  xdx+ydy+zdz=0x2+y2+z2=c2Therefore, solution is given by,x2+y2+z2=f(z2+2yz−y2)(z^2-2yz-y^2)z_x+(xy+zx)z_y= xy-zx\\ \text{By Charpit-Lagrange method :}\\ \frac{dx}{z2−2yz−y2}=\frac{dy}{xy+zx}=\frac{dz}{xy−zx}\\ \text{First characteristic equation,}\\ \frac{ dy}{xy+zx}=\frac{ dz}{xy−zx}\\\frac{ dz}{dy}=\frac{y−z}{y+z} \\ z^2+2yz−y^2=c1\\ \text{Second characteristic equation,}\\ \frac{xdx+ydy+zdz}{x(z^2-2yz-y^2)+y(xy+zx)+z(xy-zx)}=0\\\frac{xdx+ydy+zdz}{0}=0\\\implies xdx+ydy+zdz=0\\ x^2+y^2+z^2=c2 \\ \text{Therefore, solution is given by,}\\ x^2+y^2+z^2=f(z^2+2yz-y^2)


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