Differential equation---a conceptual problem

[davon806]

Homework Statement

I am reading a note on differential equation.There is a point that I don't understand,hopefully someone can explain
(Please see the attched)

Homework Equations

The Attempt at a Solution

The notes wrote " a1t + b1 x + c1 = a1T + b1X
a2t + b2t + c2 = a2T + b2X

the following sentence dX =dx + 0 and dT = dt+0
What does it mean?
I guess it was something like dX/dT = d(x+ħ)/d(t+τ)?But then I have no idea.This is the first time I have met this kind of case...
Can someone explain,thx so much

 

[Ray Vickson]

Homework Statement

I am reading a note on differential equation.There is a point that I don't understand,hopefully someone can explain
(Please see the attched)

Homework Equations

The Attempt at a Solution

The notes wrote " a1t + b1 x + c1 = a1T + b1X
a2t + b2t + c2 = a2T + b2X

the following sentence dX =dx + 0 and dT = dt+0
What does it mean?
I guess it was something like dX/dT = d(x+ħ)/d(t+τ)?But then I have no idea.This is the first time I have met this kind of case...
Can someone explain,thx so muchView attachment 90144

It says to choose $\tau$ and $\eta$ such that $\bar{x} = x + \eta$ and $\bar{t} = t + \tau$ obey the two equations you wrote above. That tells you how to find $\tau$ and $\eta$, and since the $a_i$ and $b_i$ are constants, so are $\tau$ and $\eta$. Thus, $d \bar{x} = dx + d \eta = dx$, because $d \eta = 0$ ($\eta$ is a constant).

 

[davon806]

But why does dn = 0? I have only learned dy/dx,I don't know what does it mean when they are separated(and so I post this thread :/)...

 

[Ray Vickson]

But why does dn = 0? I have only learned dy/dx,I don't know what does it mean when they are separated(and so I post this thread :/)...

You don't know why the derivative of a constant is zero? Go back to your elementary calculus notes!

 

[davon806]

Yes,I understand dy/dx = 0 if y is a constant.But the note said dy = 0,so the bottom dx is vanished,and that's make me confused

 

[Ray Vickson]

Yes,I understand dy/dx = 0 if y is a constant.But the note said dy = 0,so the bottom dx is vanished,and that's make me confused

I don't see any $y$ or $dy$ anywhere in the note.

 

[davon806]

Sorry,I mean d(something)
From the notes:
dX = dx + 0,dT = dt + 0

The problem is that I don't know what's going on because usually when you differentiate a function,say y = x^2,you write:
dy/dx = d(x^2)/dx = 2x ,the Leibniz's notation is written as dy/dx.In my notes,just like dX = dx + 0,and the "denominator" in the standard dy/dx notation vanishes,so that's why I don't understand the statement above.

 

[phyzguy]

Yes,I understand dy/dx = 0 if y is a constant.But the note said dy = 0,so the bottom dx is vanished,and that's make me confused

Writing it this way is a shorthand called implicit differentiation. You have:
\bar x = x + \eta; \frac{d \bar x}{d x} = 1 + 0
Now multiply through by dx and write:
d \bar x = dx + 0

After a bit of practice, you just skip the intermediate step and write immediately:
d \bar x = dx + 0

 

[davon806]

Thx