can u cancel infinty,,,,this is probably a very stupid/petty question!!Q is find the equation of the straight line on which the gradient of the solution curve is infinite. I made the presumption that this straight line would essentially be an isocline(hopefulli im ryt) and did the follwing,, Is is more accurate using the D.e and making bottom=0???Or is my working ok?!
dy/dx=(3y-x)/(y+x)
dy/dx=m(constant)
putting constant as infinity and rearranging to f(x)
(i) y= -x(1-m)/(m-3)
(ii) y= -x(1-inf)/(inf-3) implies(i think) y= -x(inf)/(inf)
y = -x
is this okai???
nah i think the m value should just equal infinty. If you imagine all the direction indicators on a tangent field being vertical in which case i believe there'll be no limits excuse my wording earlier. Erm my query was more in this situation by putting m = inf does this diminish the accuracy of my working when i say that y= -x(1+inf)/(inf - 3) is the equivalent of y= -x(inf/inf) in which case inf would cancel leaving me witht the right answer x + y = 0
oh ryt:P errrr well isnt it f(x)/f(x) where the bottom is only a constnt so im not reli doing that. Iim still confsed as to whether i should use limits or not?? lol i should just equate the bottom to zero