Differential form of a differential equation By multiplying the

xxgetthisar08

xxgetthisar08

Answered question

2022-03-27

Differential form of a differential equation
By multiplying the differential equation
6xydydx+x2+y2=0
by dx we see that the equation has the alternative differential form
(x2+y2)dx+6xydy=0.

Answer & Explanation

yaum3xg1

yaum3xg1

Beginner2022-03-28Added 12 answers

Step 1
When dx and dy are differential forms the equation
(x2+y2), dx +6xy, dy =0
says that along every (not necessarily closed) curve in R2 the integral of the vector field
(x2+y26xy)
must be zero. This means: if the curve is
Step 2
γ:t(γ1(t)γ2(t)),  t[0,T]
then γ(x2+y2), dx +6xy, dy =0T(γ12(t)+γ22(t))γ˙1(t)+6γ1(t)γ2(t)γ˙2(t), dt 
must be zero. In particular, if the curve is
γ:t(tγ2(t)),  t[0,T]
then γ(x2+y2),dx+6xy,dy=0Tt2+γ22(t)+6,t,γ2(t),γ˙2(t),dt (1)
must be zero. Because this holds for every T0 it follows that the integrand on the RHS must be zero. When we change notation this is nothing else than the ordinary differential equation
x2+y2+6xydydx=0 (2) we started with.
Conversely, using the same formalism it is easy to see that if y is a solution of the ODE (2) then the line integral (1) is zero along every curve. This is another way of saying that the differential form (x2+y2), dx +6xy, dy  must be zero.

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