1
$\begingroup$

I'm a student just getting started with differential equations. I saw this on a video online, and I don't understand how this equation switched forms. Maybe I'm just tired, hehe. A little help would go a long way.

$$\frac{dy}{dx} = g(x)h(y) $$

Is equivalent to

$$g(x)dx+h(y)dy = 0$$

I only got to the part where $dy/h(y) = g(x)dx$. I know I have to subtract, but I end up with $h(y)$ in the denominator.

Edit: I got it from this video at 29:20 https://www.youtube.com/watch?v=WxVaVzxsDb0

  • 0
    I don't think they're equivalent. – 2017-01-12
  • 0
    Are you sure it's not dy/dx=g(x)/h(y) and g(x)dx-h(y)dy=0? – 2017-01-12
  • 0
    I edited the post to add the video from which I got it from. – 2017-01-12

1 Answers 1

0

The guy is saying that these are two possible forms of writing separable equations, not that they are equivalent.

Notice that he wrote $G(x)$ and $H(y)$ in the second equation. This means that given $g,h$ in the first equation you can find $G,h$ such that the second equation is equivalent to the first one.

  • 0
    Thank you so much. Also, just a quick question after integrating a separable DE. Why is it that only one side gets a +c constant? – 2017-01-12
  • 0
    When you find a primitive it is never unique: all primitives are obtained by adding a constant to any given primitive (and all such functions are primitives). That's where the constant comes from. – 2017-01-12
  • 0
    Thank you very much! :) – 2017-01-12