# What is a solution to the differential equation dy/dx=e^-x/y?

Jul 16, 2016

$y = \sqrt{2 \left(C - {e}^{- x}\right)}$ and $y = - \sqrt{2 \left(C - {e}^{- x}\right)}$

#### Explanation:

This differential equation is separable, thus we only have to move things around and take integrals.

We have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{- x}}{y}$, or we can also write

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{y \cdot {e}^{x}}$

Separable differential equations require our equation to have all $y$'s and $\mathrm{dy}$'s on one side, and all $x$'s and $\mathrm{dx}$'s on the other.

In this case, we can start off by multiplying both sides by $y$.

$y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cancel{y} {e}^{x}} \cdot \cancel{y}$

$y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{e}^{x}}$

Moving our $\mathrm{dx}$ on the right by multiplying both sides the same way we get

$y \frac{\mathrm{dy}}{\cancel{\mathrm{dx}}} \cdot \cancel{\mathrm{dx}} = \frac{1}{{e}^{x}} \cdot \mathrm{dx}$

$y \cdot \mathrm{dy} = \frac{1}{{e}^{x}} \mathrm{dx}$

$y \cdot \mathrm{dy} = {e}^{- x} \mathrm{dx}$

This looks very familiar. In fact, we can integrate both sides now.

$\int y \mathrm{dy} = \int {e}^{- x} \mathrm{dx}$

$\frac{1}{2} {y}^{2} = - {e}^{- x} + C$

Our goal now is to get $y$ by itself. In order to do this, we can move a few things around again.

Multiplying both sides by $2$ yields

$\cancel{2} \cdot \frac{1}{\cancel{2}} {y}^{2} = 2 \left(- {e}^{- x} + C\right)$

${y}^{2} = 2 \left(- {e}^{- x} + C\right)$

By taking the square root of both sides we get

y = ± sqrt(2(-e^(-x) + C))

So, the general solutions to our differential equation are

$y = \sqrt{2 \left(C - {e}^{- x}\right)}$

and

$y = - \sqrt{2 \left(C - {e}^{- x}\right)}$