1. ## Implicit Solution

Would really appreciate help with this.

Show that
e^2y + e^2x = 1 defines an implicit solution of the dif-
ferential equation
e^(x-y) + e^(y-x) dy/dx = 0

I've differentiated the equation e^2y + e^2x = 1

Then tried pluging the result into the differential equation in place of dy/dx but to no avail.

Thanks!

$\displaystyle e^{x - y} + e^{y - x}\frac{dy}{dx} = 0$

$\displaystyle e^xe^{-y} + e^ye^{-x}\frac{dy}{dx} = 0$

$\displaystyle e^ye^{-x}\frac{dy}{dx} = - e^xe^{-y}$

$\displaystyle e^y\frac{dy}{dx} = - e^{2x}e^{-y}$

$\displaystyle e^{2y}\frac{dy}{dx} = -e^{2x}$

$\displaystyle \int{e^{2y}\frac{dy}{dx}\,dx} = \int{-e^{2x}\,dx}$

$\displaystyle \int{e^{2y}\,dy} = -\frac{1}{2}e^{2x} + C_1$

$\displaystyle \frac{1}{2}e^{2y} + C_2 = -\frac{1}{2}e^{2x} + C_1$

$\displaystyle \frac{1}{2}e^{2y} + \frac{1}{2}e^{2x} = C_1 - C_2$

$\displaystyle e^{2y} + e^{2x} = C$ where $\displaystyle C = 2(C_1 - C_2)$.

Any $\displaystyle C$ will satisfy the DE, so by letting $\displaystyle C = 1$ we see

$\displaystyle e^{2y} + e^{2x} = 1$ is a solution of the DE.