DE midterm problems I got wrong

[JasonJo]

Consider a DE

dy/dx = f(x,y)

Suppose that all the scaling transformations Tt, t not equal to 0, are symmetries of the DE. Show that the value of f(x,y) only depends on the ratio y/x, by showing that:

f(x,y) = f(1, y/x)

I think I have a rough idea of how to do this, my professor did it in class, but every step doesn't seem to be justified, or it's one of those solutions where the solution seems to come out of thin air, to me.

I tried:

Let T be a symmetry of the DE:
T takes (x,y) to (u = tx, v = ty)
du = t*dx, dv = t*dy

dv/du = (t/t)*(dy/dx) = dy/dx = f(y/x)

but how do you deduce that?

another problem:
Consider the 2nd order nonlinear DE:

m*(dv/dt) = mg - kv^2

at time t=0, the initial velocity is v(0) = 0.

(a) Solve for velocity v as a function of t be rewriting the above DE as a seperable first-order DE in v and t.

I don't really see how to do this, any hints?

 

[HallsofIvy]

Consider a DE

dy/dx = f(x,y)

Suppose that all the scaling transformations Tt, t not equal to 0, are symmetries of the DE. Show that the value of f(x,y) only depends on the ratio y/x, by showing that:

f(x,y) = f(1, y/x)

I think I have a rough idea of how to do this, my professor did it in class, but every step doesn't seem to be justified, or it's one of those solutions where the solution seems to come out of thin air, to me.

I tried:

Let T be a symmetry of the DE:
T takes (x,y) to (u = tx, v = ty)
du = t*dx, dv = t*dy

dv/du = (t/t)*(dy/dx) = dy/dx = f(y/x)

but how do you deduce that?

More importantly, how did f go from a function of two variables to a function of only one variable!
Try this: let v= y/x. Then y= xv so y'= xv'+ v (' means differentiation with respect to x.) f(x,y)= f(x,xv)= f(1,v) since (x,xv)-->(1,v) is a "scaling" transformation. Then differential equation becomes xv'= f(1,v)- v which is a separable equation.

another problem:
Consider the 2nd order nonlinear DE:

m*(dv/dt) = mg - kv^2

at time t=0, the initial velocity is v(0) = 0.

(a) Solve for velocity v as a function of t be rewriting the above DE as a seperable first-order DE in v and t.

I don't really see how to do this, any hints?

?? That's NOT a 2nd order equation- its first order and separable.
Separate it as m (dv/(mg- kv^2))= dt and use partial fractions to integrate the right side.

 

[JasonJo]

can i also scale by t = 1/x

ie, let T be the transformation: (x y) |--> (x*t y*t) and let t = 1/x?? since all scaling transformations are symmetries of the DE?

 

[JasonJo]

bump


anything guys?

 

[AKG]

You need to show that all scalings are symmetries, not that all symmetries are scalings. Take a scaling, apply it to your DE, and prove that the resulting thing is the same as what you started with (I'm assuming that's what you mean by a symmetry).

dv/du = (t/t)*(dy/dx) = dy/dx = f(y/x)

You mean dv/du = (t/t)*(dy/dx) = dy/dx = f(x,y), since dy/dx = f(x,y) by the original equation. Anyways:

dv/du = dv/dy * dy/dx * dx/du = t * dy/dx * 1/(du/dx) = t * dy/dx * 1/t = (t/t)*(dy/dx)