Question

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

Answer 1

`y = sqrt(2(C-e^(-x)))` and `y = - sqrt(2(C-e^(-x)))`

Explanation:

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

We have `(dy)/(dx) = (e^(-x))/(y)`, or we can also write

`(dy)/(dx) = (1)/(y*e^(x))`

Separable differential equations require our equation to have all `y`'s and `dy`'s on one side, and all `x`'s and `dx`'s on the other.

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

`y(dy)/(dx) = (1)/(cancel(y) e^(x)) * cancel(y)`

`y(dy)/(dx) = (1)/(e^(x))`

Moving our `dx` on the right by multiplying both sides the same way we get

`y(dy)/cancel(dx) * cancel(dx)= (1)/(e^(x)) * dx`

`y* dy = (1)/(e^(x)) dx`

`y* dy = e^(-x) dx`

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

`int ydy = int e^(-x) dx`

`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) * 1/cancel(2) y^2 = 2(-e^(-x) + C)`

`y^2 = 2(-e^(-x) + C)`

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(C-e^(-x)))`

and

`y = - sqrt(2(C-e^(-x)))`