Answer to Question #171177 in Differential Equations for melvin

Question

(y-2)dx-(x-y-1)dy=0

Answer 1

(y-2)dx (x-y-1) dy =0

we get

dy/dx=(y-2)/(x-y-1)

let x=u+p; and y=v+r;

dx=du; dy=dv;

dy/dx=[(v+r)-2]/[(u+p)-(v+r)-1]

We get

dv/du[v+(r-2)]/[u-v+(p-r-1)]

thus

r-2=0; hence r=2;

p-r-1=0; hence substituting we get p=3;

x=u+3; y=v+2 ; which becomes

dv/du=v/(u-v);

let v/u =z; implies v=uz

dv/du=z+u (dz/du)

z+u(dz/du)=z/(1-z)

$\intop$ (z^-2-z^-1)=$\intop$ du/u

ln uz=-(1/z)+d

Solve the values of u,v, z, p and r

(y-2)=Ce^{-[(x-3)/(y-2)]}