Integrate dy/dx=0 and arbitrary constant?

[DryRun]

Homework Statement
Integrate:
\frac{dy}{dx}=0

The attempt at a solution
\int \frac{dy}{dx}=\int 0\,.dx
The answer is: y=0 or y=0+A?
This is the part which is confusing, as i know that integral of 0 is 0, but do i have to add a constant of integration??

 

[dikmikkel]

How about a dy/dx which is as positive as negative over the interval, that could be zero. fix it was e.g. sin(x)
But yeah, a constant.

 

[DryRun]

Now that i think about it, here is a simple example:
Let y=5
\frac{dy}{dx}=0
Now, if i integrate 0, then i should normally get 5. So, the constant of integration has to be involved.

 

[Mark44]

Homework Statement
Integrate:
\frac{dy}{dx}=0

The attempt at a solution
\int \frac{dy}{dx}=\int 0\,.dx

The above should be
\int \frac{dy}{dx}~dx=\int 0\,.dx
You are integrating both sides of the original equation, with respect to x.

The answer is: y=0 or y=0+A?
This is the part which is confusing, as i know that integral of 0 is 0, but do i have to add a constant of integration??

Zero is merely one antiderivative of 0.

A simpler approach is as follows.
Since dy/dx = 0, then y must be a constant. IOW y $\equiv$ C.

Also, when you integrate both sides of an equation, there is a constant of integration for each side. So the integration that you did would look like this:
y + C₁ = 0 + C₂

Of course, you can subtract C₁ from both sides to end up with y = C, where C = C₂ - C₁.

 

[DryRun]

Thank you for the clarification, Mark44.

 

[HallsofIvy]

Most teachers will mark off on homework or a test if you do not include the "constant of integration". If F(x) is an anti- derivative of f(x)- that is, if F'(x)= f(x), then \int f(x)dx= F(x)+ C where C can be any number.