e^(xy) + y^2 = sin(x + y)

asked Dec 9, 2016

It's unlikely that y can be expressed in terms of x but we can make some observations.

The expression on the right has a range of -1 to 1, but the expression on the left is always strictly positive. Therefore we know that 0<sin(x+y)≤1 and 0<e^(xy)+y^2≤1.

When y=0, x=(4n+1)π/2 where n is an integer, because sin(x)=1 when y=0.

So for one value of y there are multiple values of x.

When x=0, y^2=sin(y)-1, no solutions for y. When x<0 and y>0, y≤1-e^(xy) so y<1 (radians).

When e^(xy)<1, xy<0 and y>0, so 0≤y<1 and x<0 (or -1<y≤0 and x>0).

answered Dec 18, 2016 by Top Rated User (429,280 points)
Oh, so you finaly answer my hardest question eh? Well you earn yourself a best answer for that.
Well, I didn't solve it, but I'm still looking into it and building a picture.
ITS HARD AND TOUGH HUH?

It may not be possible, but what is possible is to graph the relationship, which is what I'm trying to do accurately.

I didn't need help with this. I wanted to see if you could solve this because your such a "smarty pants", so I looked up the hardest algebra in the world that no one can't answer! NOT EVEN YOU!