First order differential equation

[squareroot]

Homework Statement

I'm starting college this autumn(physics) and I started learning some calculus on my own, basic stuff like first order differential equation and so on.Recently i stumbled on something that i don t understand.I was reading the course and re-solving the given examples for myself.So you have the equation

\frac{dy}{dx} = \frac{ \frac{y}{x} - 1 }{ \frac{y}{x} + 1 }

Homework Equations

calculus

The Attempt at a Solution

Given that the equation is already solved but i didn't understand one of the steps if the solutions i'll show you where i get stuck.

So you substitute y=tx and here comes the boom, by doing this substitution they say the result is
\frac{t+1}{t²+1}dt =-\frac{dx}{x} and from here on it s quite simple, you just integrate both sides and get the result.But i didn't understand the steps from substituting y=xt to getting to that result.I ll show you how i tried to solve it but got to a different result.

So if y=tx --> dy/dx=tdx and it s clear that if you substitute dy/dx in that equation you don t get their answer.

My main problem is that, in high school we didn't use this notation, we used f'(x) instead of dy/dx and so on, so I'm having a bit of trouble getting used with this notation, but these are the one used in physics so now i have to squash my brains a little and try to get used to them.

If someone can explain that step to me and maybe give me a few tips on how to make the transition between these two notation easier it would be great!

Thank you

 

[HallsofIvy]

"Here comes the boom"?

Seeing the "y/x" on the right, I think it would be obvious to set t= y/x which is the same as y= xt. Differentiating both sides with respect to x,
\frac{dy}{dx}= x\frac{dt}{dx}+ t.
And the whole point of that substitution is that
\frac{\frac{y}{x}- 1}{\frac{y}{x}+ 1}= \frac{t- 1}{t+ 1}
So the equation becomes
x\frac{dt}{dx}+ t= \frac{t- 1}{t+ 1}
x\frac{dt}{dx}= \frac{t-1}{t+ 1}- t= \frac{t- 1}{t+ 1}- \frac{t(t+ 1)}{t+1} Combine the fractions on the right and you have a relatively simple "separable" equation.

(Personally, I thing dy/dx is a better notation than y' but you can translate back:
y'= (y/x- 1)/(y/x+ 1). Let t= y/x so that y= xt and y'= xt'+ t. The equation becomes xt'+ t= (t- 1)/(t+ 1). xt'= (t- 1)/(t+ 1)- t= (t- 1)/(t+ 1)- (t(t+ 1))/(t+ 1). Of course, when you learned integration, you must have learned that if y'= f(x) then y= \int f(x)dx which follows more easily from dy/dx= f(x), dy= f(x)dx.)

 

[squareroot]

Ah... Got it! thank you