# DE#3

#### idontknow

$$\displaystyle dx+(e^y -x)dy=0$$.

#### topsquark

Math Team
Just out of curiousity, where are you getting these DEqs from?

-Dan

#### idontknow

Just out of curiousity, where are you getting these DEqs from?

-Dan
from a very old book , it is an exact DE.

• topsquark

#### topsquark

Math Team
Well, if it's exact then you have
$$\displaystyle 1 + \left ( e^y - x \right ) \dfrac{dy}{dx} = 0$$

and you have the general form
$$\displaystyle M(x, y) + N(x, y) \dfrac{dy}{dx} = 0$$

then what are M(x, y) and N(x, y) and how do you use them?

Let us know what you've got and we'll go from there. The solution method isn't particularly hard but it does require some intuition.

-Dan

• idontknow

#### skipjack

Forum Staff
The equation (which isn't exact) implies $$\displaystyle e^{-y}\frac{dx}{dy} - xe^{-y} = -1$$.
Integrating with respect to y gives $e^{-y}x = -y + \text{C}$, where $\text{C}$ is a constant,
so $x = (\text{C} - y)e^y$.

• topsquark and idontknow

#### topsquark

Math Team
The equation (which isn't exact) implies $$\displaystyle e^{-y}\frac{dx}{dy} - xe^{-y} = -1$$.
Integrating with respect to y gives $e^{-y}x = -y + \text{C}$, where $\text{C}$ is a constant,
so $x = (\text{C} - y)e^y$.
Nice!

-Dan