Re: How to Plot this equation ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg47121] Re: How to Plot this equation ?*From*: drbob at bigfoot.com (Bobby R. Treat)*Date*: Fri, 26 Mar 2004 03:56:23 -0500 (EST)*References*: <c3uekb$9u0$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Solving for y doesn't work, but we can solve for x and transpose the resulting graph. x in terms of y involves ProductLog, which has a discontinuity at -1/E, so I locate the y values that get us to that value, then restrict the function to keep it "real". After drawing x as a function of y, I reverse points in the Line directive to transpose the graph. eqn = x*Exp[x*y + 0.8] + Exp[y^2] == 3; Off[InverseFunction::ifun, Solve::ifun] inverse[y_] = x /. First[Solve[eqn, x]] yLimits = (y /. FindRoot[Cases[inverse[y], ProductLog[a_] :> a == -1/ E], {y, #1}] &) /@ {0, 1} invPlot = Plot[inverse[y], Evaluate[{y, Sequence @@ yLimits}]]; Show[invPlot /. Line[a_] :> Line[Reverse /@ a]]; (1/y)*ProductLog[ 9.080790118821775*^-9* (1.48443789*^8 - 4.9481263*^7*E^y^2)*y] {-0.4657975933929999, 1.1455559242943212} Bobby m004202002 at yahoo.com (why) wrote in message news:<c3uekb$9u0$1 at smc.vnet.net>... > Hi, > > How I can Plot tis equation with mathematica? > > x exp(xy+0.8) + exp(y^2)=3 > > Please help me . > > Thank You .