Question

Solve y'=(-x+3y-14)/(x+y-2) implicitly.

Answer 1

@SithsAndGiggles

Answer 2

Have you tried applying that method we discussed last time?

Answer 3

u=x+h
v=y+k
dy/dx=dv/du
dv/du=(-(u-h)+3(v-k)-14)/((u-h)+(v-k)-2)
dv/du=(-u+3v+(h-3k-14))/(u+v+(-h-k-2))
h-3k-14=0
-h-k-2=0
k=-4
h=2
dv/du=(-u+3v)/(u+v)
dv/du=(-1+3v/u)/(1+v/u)
v=uz
dv/du=u*dz/du+z
d*dz/du+z=(-1+3z)/(1+z)
u*dz/du=(-1+3z)/(1+z)-z
u*dz/du=(-1+3z)/(1+z)-(z(1+z))/(1+z)
=(-1+3z-z-z^2)/(1+z)
=(-1+2z-z^2)/(1+z)
=(z^2-2z+1)/(z+1)
=(z-1)^2/(z+1)
(z+1)/(z-1)^2 dz=du/u
u=z-1
du=dz
z+1=u+2
ln abs(z-1)-2/(z-1)
ln abs(z-1)-2/(z-1)=ln abs(u)+C
ln abs(z-1)-ln abs(u)-2/(z-1)=C
ln abs((z-1)/u)-2/(z-1)=C
am I right so far?

Answer 4

Thank you, sithandgiggles.