What is the difference between these types of differential equations?

[Jhenrique]

Which the difference between diff equations of kind: $$\frac{dy}{dx} = \exp(x)$$ $$\frac{dy}{dx} = 1/x$$
and diff equations of kind:
$$\frac{dy}{dx} = y$$ $$\frac{dy}{dx} = \frac{1}{\exp(y)}$$ ?

 

[Simon Bridge]

Nothing.
Well... apart from the obvious.

You can call the variables anything you like.
Try solving them and see what you get.

 

[Jhenrique]

A general linear diff equation of 1st order have a form: $$f(x)y'(x)+g(x)y(x)=h(x)$$ The 3 first equations fits into this definition, but the last no. However, the last equation is linear too. This is a contradiction?

 

[justawebuser]

Nothing,
As already mentioned both are the same.
What do you mean by that?

 

[justawebuser]

A general linear diff equation of 1st order have a form: $$f(x)y'(x)+g(x)y(x)=h(x)$$ The 3 first equations fits into this definition, but the last no. However, the last equation is linear too. This is a contradiction?

Note that when you want solve an equation you must specify the independent variable , first tell us what variable your are assuming as independent one,

 

[Mark44]

A general linear diff equation of 1st order have a form: $$f(x)y'(x)+g(x)y(x)=h(x)$$ The 3 first equations fits into this definition, but the last no. However, the last equation is linear too. This is a contradiction?

No, the last equation is not linear. In a linear differential equation y, y', y'', ... occur to the first power only, but can be multiplied by functions of the independent variable. With the right side being e^-y, the equation is not linear.

 

[Mark44]

A general linear diff equation of 1st order have a form: $$f(x)y'(x)+g(x)y(x)=h(x)$$ The 3 first equations fits into this definition, but the last no. However, the last equation is linear too. This is a contradiction?

Note that when you want solve an equation you must specify the independent variable , first tell us what variable your are assuming as independent one,

I'm assuming that the independent variable is x, and y is the dependent variable. That seems fairly clear from what Jhenrique wrote.

 

[Jhenrique]

No, the last equation is not linear. In a linear differential equation y, y', y'', ... occur to the first power only, but can be multiplied by functions of the independent variable. With the right side being e^-y, the equation is not linear.

But the antiderivative's solution of $\frac{dy}{dx} = \frac{1}{x}$ is log(x) and the antiderivative's solution of $\frac{dy}{dx} = \frac{1}{\exp(y)}$ is log(x) too. So why the first is linear and the second no?

 

[Mark44]

But the antiderivative's solution of $\frac{dy}{dx} = \frac{1}{x}$ is log(x) and the antiderivative's solution of $\frac{dy}{dx} = \frac{1}{\exp(y)}$ is log(x) too. So why the first is linear and the second no?

Two things:
1. The solution of a differential equation has nothing to do with whether the equation is linear or not.
2. The two equations do NOT have the same solution.

$\frac{dy}{dx} = e^{-y}$
$\Rightarrow e^y dy = dx$
$\Rightarrow \int e^y dy = \int dx$
$\Rightarrow e^y = x + C$
$\Rightarrow y = ln(x + C)$
This solution is different from the one for dy/dx = 1/x.