What is the General Solution to the Differential Equation?

[llorgos]

Hi! I would like to ask what is the general solution of the following differential equation
\frac{\partial X_x}{\partial t} = - \frac{\partial X_t}{\partial x}

Thank you very much.

P.S. If you have some good resiource about this tyoe of equation to recommend please do so.

 

[CompuChip]

What is X_x - does that stand for \frac{\partial X}{\partial x} ?
Because then you are basically asking about X_{xt} = -X_{tx}.

In that case you should be looking at "weird" functions - given Clairaut's theorem at least the partial derivatives should not be continuous.

 

[llorgos]

Hi.

If you want I can write it as \frac{\partial X}{\partial t} = -\frac{\partial T}{\partial x} where T = T(x,t) and X = X(x,t) in general.

I know they must be equal to a constant. Please correct me if I am wrong.

Thank you.

 

[Office_Shredder]

So the functions on the left and right hand side are not equal in general? And you're asking what the general form for X and T is as separate functions?

 

[JJacquelin]

\frac{\partial X}{\partial t} = -\frac{\partial X}{\partial x}
X=f(t-x) any derivable function f

 

[HallsofIvy]

Hi.

If you want I can write it as \frac{\partial X}{\partial t} = -\frac{\partial T}{\partial x} where T = T(x,t) and X = X(x,t) in general.

I know they must be equal to a constant. Please correct me if I am wrong.

Thank you.

Let X(x, t)= f(x, t) be any differentiable function of x and t and define T(x, t)= -f(t, x).
For example, take X(x, t)= x+ t^2, T(x, t)= -t- x^2. Then \partial X/\partial x= 1= -\partial T/\partial t.

I don't know what you mean by "they must be equal to a constant".