Question

Reduce each of the following equations into canonical form and find the general solution: (a) Uz + xy = u, (b) ux + x +y = y, (C) ux + 2xy uy = x, (d) Uz – yuy – u = 1.

Answer 1

c)

$p+2xyq=x$

$\frac{dx}{1}=\frac{dy}{2xy}=\frac{dz}{x}$

$2xdx=dy/y$

$x^2=lny+c_1$

$xdx=dz$

$x^2=2z+c_2$

$F(c_1,c_2)=F(x^2-lny,x^2-2z)=0$

b)

$\frac{dx}{1}=-\frac{du}{x}=\frac{dy}{0}$

$xdx=-du$

$x^2=-2u+c_1$

$x^2+2u=c_1$

$y=c_2$

$F(c_1,c_2)=F(x^2+2u,y)=0$

d)

$\frac{dz}{1}=-\frac{dy}{y}=\frac{du}{u+1}=\frac{dx}{0}$

$z=-lny+lnc_1$

$e^zy=c_1$

$-lny=ln(u+1)-lnc_2$

$c_2=y(u+1)$

$x=c_3$

$F(c_1,c_2,c_3)=F(e^zy,y(u+1),x)=0$

a)

$\frac{dz}{1}=\frac{du}{u-xy}=\frac{dx}{0}=\frac{dy}{0}$

$x=c_1$

$y=c_2$

$dz=du/(u-c_1c_2)$

$z=ln(u-c_1c_2)+lnc_3$

$e^z=c_3(u-c_1c_2)=c_3(u-xy)$

$e^z/(u-xy)=c_3$

$F(c_1,c_2,c_3)=F(x,y,e^z/(u-xy))=0$