
I'm hitting a brick wall with $\dfrac{dx}{ds} = 1-y^2$. Is it possible for me to let $c = 1-y^2$?
Would appreciate any help.

I'm hitting a brick wall with $\dfrac{dx}{ds} = 1-y^2$. Is it possible for me to let $c = 1-y^2$?
Would appreciate any help.
You can use separation of variables to solve $\frac{dx}{ds} = 1-y^2 $ as
$$ x= (1-y^2 )s+k. $$
On my solution it says: u(x,y,0)= e^(y-x0)^(2) then, u(x,y,t) = e^(y-(x+(y-1)^(2)t)^(2)) which I clearly have some problem wrapping my head around. If you would help me.
– guest Mar 14 '15 at 06:14