Question #289453

Find the integral surface of the equation:


(x²-yz)p+(y²-zx)q =z²-xy


passing through the line x=1 , y=0

1
Expert's answer
2022-01-27T00:46:57-0500

To find the integral surface of the given equation, we proceed as follows:\text{To find the integral surface of the given equation, we proceed as follows:}

The auxiliary equations are;

dxx2yz=dyy2zx=dzz2xy\displaystyle \frac{dx}{x^2-yz}=\frac{dy}{y^2-zx}=\frac{dz}{z^2-xy}\\

dxdyx2yzy2+zx=dydzy2zxz2+xy=dzdxz2xyx2+yzdxdy(xy)(x+y+z)=dydz(x+y+z)(yz)=dzdx(x+y+z)(zx)dxdyxy=dydzyz=dzdxzx(1)\displaystyle \Rightarrow \frac{dx-dy}{x^2-yz-y^2+zx}=\frac{dy-dz}{y^2-zx-z^2+xy}=\frac{dz-dx}{z^2-xy-x^2+yz}\\ \Rightarrow \frac{dx-dy}{(x-y)(x+y+z)}=\frac{dy-dz}{(x+y+z)(y-z)}=\frac{dz-dx}{(x+y+z)(z-x)}\\ \Rightarrow \frac{dx-dy}{x-y}=\frac{dy-dz}{y-z}=\frac{dz-dx}{z-x}\cdots\cdots\cdots\cdots(1)\\

Integrating first two members of (1) yields;

log(xy)=log(yz)+logc1logxyyz=logc1xyyz=c1\displaystyle \log(x-y)=\log(y-z)+\log c_1\\ \Rightarrow\log \frac{x-y}{y-z}=\log c_1\\ \Rightarrow \frac{x-y}{y-z}=c_1\\

Similarly from the last two members of (1), we have;

yzzx=c2\displaystyle \frac{y-z}{z-x}=c_2

The required solution is;

f[xyyz,yzzx]=0\displaystyle f\left[\frac{x-y}{y-z},\frac{y-z}{z-x}\right]=0


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