Recent content by inglezakis

Thats a good answer indeed.

The constraint in the example we discuss is, xy=a, so f(x)=y=a/x and thus it is a function. What you mean by arbitrary function? Lets say you don't eliminate y and you use the constraint to solve the equation: ydx+ydy=0, constaint: xy=a By substituting x with a/y and after some...

Hi JJacquelin, i don't think the supposition is wrong. You can still solve the equation and get the result. By the way, this equation represents a physical system (fluid flow) and i know that is true and the solution is correct. The target of the question is to find out what happens to a...

The point of my question is that when we divide a differential equation by a function or variable we result in different solution (not always). Take the example: ydx+ydy=0, constaint: xy=a By substituting x with y/a and after some manipulations we arrive to (-a/y)dy+ydy=0 and on integration...

And why the supposition is wrong?

Consider the differential equation dx+ydy=0, the integration leads to (x2-x1)+(y2^2-y1^2)/2=0 (1) Suppose we know that y/x = const. Lest proceed to the following manipulation on the initial equation, by dividing by (x), then dx/x+(y/x)dy=0, now the integration gives...