Solving the Equation: (-y/x^2)dx+(1/x)dy=d(y/x)

[asdf1]

why does (-y/x^2)dx+(1/x)dy=d(y/x)?

 

[lurflurf]

because
df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy
and
\frac{\partial}{\partial x}(\frac{y}{x})=\frac{-y}{x^2}
and
\frac{\partial}{\partial y}(\frac{y}{x})=\frac{1}{x}

 

[asdf1]

but how do you think of going from the left side to the right side if you haven't seen the right side before?

 

[asdf1]

it's hard to understand...

 

[Astronuc]

\frac{d}{dx}(\frac{y}{x})= \frac{d}{dx} (y*\frac{1}{x})

If one has done the derivative of a product, then one knows -

\frac{d}{dx}(u(x)*v(x))= v(x) * \frac{d}{dx} u(x) + u(x) * \frac{d}{dx} v(x)

This can be verified with the definition of the derivative.

Then \frac{d}{dx}(\frac{1}{x})= -\frac{1}{x^2},

which can also be shown using the definition of the derivative.

 

[matt grime]

it's hard to understand...

no it isn't, it's a definition. if you don't know what the definitions are then of course it will seem hard, but this is exactly what the notation means.

 

[HallsofIvy]

Yes, it is clear that d\frac{y}{x}= -\frac{y}{x^2}dx+ \frac{1}{x}dx because \frac{d(\frac{a}{x})}{dx}= -\frac{a}{x^2} for any constant a and \frac{d(\frac{y}{a})}{dy}= \frac{1}{a} for any constant a.

But how would you go from -\frac{y}{x^2}dx+ \frac{1}{x}dx to d{(\frac{y}{x})}? That, I think, is the question being asked and I'll bet it's in your textbook: look under "exact differentials" or "integrals independent of path". (Actually, going that direction gives a more general answer- you have a "constant of integration".)

You know that dF(x,y)= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y} dy. So whatever F is, if dF= -\frac{y}{x^2}dx+ \frac{1}{x}dx then we must have \frac{\partial F}{\partial x}= -\frac{y}{x^2} and \frac{\partial F}{\partial y}= \frac{1}{x}.

In particular, taking the anti-derivative of \frac{\partial F}{\partial x}= -\frac{y}{x^2} , we have (remembering that the partial derivative treats y like a constant), F(x,y)= \frac{y}{x}+ g(y) because the anti-derivative of \frac{a}{x^2}= ax^{-2} is -ax^-1 and here a is -y. Notice that g(y)! That's the "constant" of integration- no matter what function of y g is, taking the partial derivative with respect to x is 0.

Now, differentiate that with respect to y: \frac{\partial \frac{y}{x}+ g(y)}{\partial y}= \frac{1}{x}+ g'(y) (Since g is a function of y only, the derivative of g is an ordinary derivative). Now compare that with \frac{\partial F}{\partial y}= \frac{1}{x}. We must have \frac{1}{x}+ g'(y)= \frac{1}{x} which just tells us that g'(y)= 0 so g(y) really is a constant!

F(x,y)= \frac{y}{x}+ C or d(\frac{y}{x}+C)= -\frac{y}{x^2}dx+ \frac{1}{x}dx. Of course, your example is taking C= 0.

 

[asdf1]

thank you! :) that clears a lot of things up~