# Domain of $e^{2y} + e^{2x} = 1$ as implicit solution to ODE

Paraphrasing a problem from Ordinary Differential Equations by Tenenbaum and Pollard,

Determine whether the equation $e^{2y} + e^{2x} = 1$ defines $y$ as an implicit function of $x$. If so, determine whether it is an implicit solution of the differential equation $e^{x - y} + e^{y - x}\frac{dy}{dx} = 0$.

I solved for $y$ by subtracting $e^{2x}$ from both sides and taking the natural logarithm to get $$y = \frac{1}{2}\ln(1 - e^{2x})$$ which defines a function for $1 - e^{2x} > 0$, or $x < 0$. Using implicit differentiation, I found that an implicit function from the given equation will solve the ODE. Combined with the implicit function, the equation is an implicit solution for $x < 0$.

The book says that it implicitly solves the ODE for $x \neq 0$. How did $x > 0$ get picked up? I see that WolframAlpha gives the same domain, but with a complex function.

• You have not given the ODE in full Jan 6 '16 at 6:44
• @Daniel: Fixed; I don't remember the equation being that implicit. Jan 6 '16 at 6:50

For $x>0$ note that \begin{equation*} e^{i\pi +2y}=e^{2x}-1 \end{equation*} and \begin{equation*} y=-\frac{i\pi }{2}+\ln (e^{2x}-1) \end{equation*}
• Is there anyway to get $x > 0$ without using Euler's formula? I assumed that the text was dealing with real functions (although I don't remember it saying that). Jan 6 '16 at 17:23
• 1) Note that for $x>0$ $exp[2y]$ is non-postive so already here you are confronted with a complex number in the argument of the exponential. If you want to stick to real numbers the answer is that there are no solutions for $x>0$. Note also that $dy/dx$ remains real in this case as well. Jan 7 '16 at 9:02
• @Urgje: did you mean for $x>0, exp[i\pi +2y]$ is non-positive?