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

Question

Christopher Allers · Accepted Answer

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

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)))#

Charles Arocha · Answer

undefined To solve the differential equation $\frac{dy}{dx} = \frac{e^{-x}}{y}$, we can separate the variables and integrate both sides.

Rearrange the equation to separate the variables:
$$y dy = e^{-x} dx$$

Integrate both sides:
$$\int y dy = \int e^{-x} dx$$

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

Here, $C$ is the constant of integration. Therefore, the solution to the differential equation is:
$$y^2 = -2e^{-x} + C'$$

Where $C'$ is a constant that encompasses $2C$, which can be determined based on initial conditions or additional information provided.