Null Curves of Linear Differentials

[Spriteling]

Homework Statement

a = (y^3 + y)dx + (xy^2 + x)dy = A1dx + A2dy
Characterise the set of points in R2 which can be joined to (1,1) by a null curve.

Homework Equations

If v = z(t) is a piecewise continuous curve, and dv/dt lies in the null space of [A1(x(t)),A2(x(t))] then it is a null curve.

The Attempt at a Solution

I guess my problem with this is that I don't know how to go about finding the different z(t). The book I'm working from does an example for the differential a = dy + xdz but it's fairly simplistic and I'm having trouble actually applying it to this example. I guess I should try to compute da but I'm still a bit confused on how to do that, as I've solely been working for a textbook without very many examples...

 

[HallsofIvy]

A "null curve" for such a differential is a curve on which it is equal to 0 so you are trying to solve the differential equation (y^3 + y)dx + (xy^2 + x)dy = 0 subject to the initial condition that y(1)= 1.

Since xy^2+ x= x(y^2+ 1), that equation is separable. It is the same as
\frac{y^2+ 1}{y^3+ y}dy= \frac{dx}{x}
Just integrate both sides.

 

[Spriteling]

A "null curve" for such a differential is a curve on which it is equal to 0 so you are trying to solve the differential equation (y^3 + y)dx + (xy^2 + x)dy = 0 subject to the initial condition that y(1)= 1.

Since xy^2+ x= x(y^2+ 1), that equation is separable. It is the same as
\frac{y^2+ 1}{y^3+ y}dy= \frac{dx}{x}
Just integrate both sides.

Shouldn't it be
\frac{y^2+ 1}{y^3+ y}dy= \frac{-dx}{x} ?

Other than that, thanks! That makes sense! :)