Find the integral surface of quasi linear partial differential equation xzp+yzq=-xy which passes through the curve y=x²,z=x³
xzp+yzq=−xyThis is lagrange’s equationPp+Qq=Rdxp=dyQ=dz−xydxxz=dyyz ⟹ dxx=dyy ⟹ ∫dxx=∫dyyInx=Iny+Ina ⟹ Inxy=Inaxy=aydx+xdy−2zdzxyz+xyz−2xyz=ydx+xdy−2zdz0d(xy)−2zdz=0xy−z2=bψ(xy,xy−z2)=0.xzp+yzq=-xy \\ \text{This is lagrange's equation}\\ Pp+Qq=R\\ \frac{dx}{p}=\frac{dy}{Q}=\frac{dz}{-xy}\\ \frac{dx}{xz}=\frac{dy}{yz} \implies \frac{dx}{x}=\frac{dy}{y}\\ \implies\int\frac{dx}{x}=\int\frac{dy}{y}\\ Inx=Iny+Ina\\ \implies In\frac{x}{y}=Ina\\ \frac{x}{y}=a\\ \frac{ydx+xdy-2zdz}{xyz+xyz-2xyz}=\frac{ydx+xdy-2zdz}{0}\\ d(xy)-2zdz=0\\ xy-z^2=b\\ \psi(\frac{x}{y},xy-z^2)=0.xzp+yzq=−xyThis is lagrange’s equationPp+Qq=Rpdx=Qdy=−xydzxzdx=yzdy⟹xdx=ydy⟹∫xdx=∫ydyInx=Iny+Ina⟹Inyx=Inayx=axyz+xyz−2xyzydx+xdy−2zdz=0ydx+xdy−2zdzd(xy)−2zdz=0xy−z2=bψ(yx,xy−z2)=0.
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