Did I do this right? (Differential Eqtns)

[Color_of_Cyan]

Homework Statement

Show that the given equation is an implicit solution of the given differential equation:

Homework Equations

e^{(x - y)} + e^{(y - x)} (dy / dx) = 0

e^2y + e^2x = 1

The Attempt at a Solution

(e^2y) = 1 - e^2x

solve for dy/dx
dy/dx (e^2y) = dy/dx (1 - e^2x)

(2e^2y)(dy / dx) = -2e^2x

dy / dx = -e^2x / e^2y

Substitute: e^{(x - y)} + { e^{(y - x)} * [ -e^2x / e^2y ] } = 0

Now I am uncomfortable on my algebra here, and I try to muliply out the

{ e^{(y - x)} * [ -e^2x / e^2y ] }

(please do not laugh if I am wrong) :

e^{(x - y)} -[/color] e^{(y - x + 2x - 2y)} = 0

e^{(x - y)} - e^{(x - y)} = 0

 

[Stephen Tashi]

I think you algebra is OK.

I don't like your notation for differentiation.

Does you class actually show differentiation on both sides of the equation as:
dy/dx (e^2y) = dy/dx (1 - e^2x) ?

I prefer something like

D_x ( e^{2y}) = D_x (1 - e^{2x})

or

\frac {d( e^{2y}) }{dx} = \frac { d(1-e^{2x})}{dx}


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

 

[Mark44]

To add to what Stephen Tashi said about your differentiation notation, dy/dx means the derivative of y with respect to x. An expression such as
dy/dx(e^2y) means the derivative of y with respect to x times e^2y. It does NOT mean that you are taking the derivative of e^2y.

It's important to understand the difference between dy/dx(e^2y) and d/dx(e^2y). The first is a product of two functions; the second indicates an operation that hasn't been performed yet.