# Seperable Differential Equation

mathor345

## Homework Statement

$$\frac{dy}{dx} = e^{3x-3y}$$

## Homework Equations

$$\int e^{u}du = e^u + C$$

## The Attempt at a Solution

$$\frac{dy}{dx} = \frac{e^{3x}}{e^{3y}}$$

$$e^{3y} dy = e^{3x} dx$$

$$ln (e^{3y}) dy = ln (e^{3x}) dx$$

$$3y dy = 3x dx$$

Integrate...

As you can see I'm doomed to get x = y

Last edited:

Mentor

## Homework Statement

$$\frac{dy}{dx} = e^{3x-3y}$$

## Homework Equations

$$\int e^{u}du = e^u + C$$

## The Attempt at a Solution

$$\frac{dy}{dx} = \frac{e^{3x}}{e^{3y}}$$

$$e^{3y} dy = e^{3x} dx$$
OK to here. When you integrate, don't forget the constant of integration.

$$ln (e^{3y}) dy = ln (e^{3x}) dx$$

Gold Member
$$e^{3y} dy = e^{3x} dx$$

$$ln (e^{3y}) dy = ln (e^{3x}) dx$$

This last step doesn't work. If you tried to take the log of both sides you would get:
$$\ln{(e^{3y} dy)} = \ln{(e^{3x} dx)}$$
which doesn't make sense. Try integrating both sides instead.

Homework Helper
Gold Member
Look at the last but one line. d/dy of what gives you e3y ?

mathor345
This last step doesn't work. If you tried to take the log of both sides you would get:
$$\ln{(e^{3y} dy)} = \ln{(e^{3x} dx)}$$
which doesn't make sense. Try integrating both sides instead.

$$\int e^{3y} dy = \int e^{3x} dx$$

$$\frac{e^{3x}}{3} = \frac{e^{3y}}{3}$$

$$e^{3x} + C = e^{3y} + C$$

$$e^{3x} + C = e^{3y}$$

$$ln (e^{3x} + C) = ln (e^{3y})$$

$$ln (e^{3x} + C) = 3y$$

$$\frac{1}{3} ln (e^{3x} + C) = y$$

... I think this is the right answer, but it could very well be

$$\frac{1}{3} ln (e^{3y} + C) = x$$

Depending on where you put the constant of integration?

Last edited:
Mentor
You forgot the constant of integration...

Gold Member
Both answers are correct. Changing where you put the 'C' just changes the value of 'C', not the form of the solution, but the first form is preferable because it gives an explicit function for y.

(If you ever want to know whether your answer is right, just plug it back in to the original differential equation and see if it works)

mathor345
Thanks for the help everyone.

Mentor
$$\int e^{3y} dy = \int e^{3x} dx$$

$$\frac{e^{3x}}{3} = \frac{e^{3y}}{3}$$
The above should be
$$\frac{e^{3x}}{3} = \frac{e^{3y}}{3} + C$$

$$e^{3x} + C = e^{3y} + C$$
You're not going to have the same constant on both sides. The best way around having to work with two different constants is to use a single constant on one side, usually the right side.
$$e^{3x} = e^{3y} + C$$

$$e^{3x} + C = e^{3y}$$
Or you could do it like you did above. That's fine, too.
$$ln (e^{3x} + C) = ln (e^{3y})$$

$$ln (e^{3x} + C) = 3y$$

$$\frac{1}{3} ln (e^{3x} + C) = y$$

... I think this is the right answer, but it could very well be

$$\frac{1}{3} ln (e^{3y} + C) = x$$

Depending on where you put the constant of integration?