An ODE (Ordinary Differential Equation) of order n becomes a relation: F ( x ,...



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An ODE (Ordinary Differential Equation) of order n becomes a relation:
F ( x , y , y ( 1 ) , . . . , y ( n ) ) = 0
Then F ( x , y , y ( 1 ) ) = 0 defines an ODE of order one. In "basic standard texts", for purposes of simplicity, is assumed that some ODE of first order can take the form:
y ( 1 ) = f ( x , y )
for certain suitable f. Here is my "silly" question: What if that assumpion is not possible?
For example how I can deal with equations of the form:
( y ( 1 ) ) 5 + s e n ( y ( 1 ) ) + e y ( 1 ) + x = 0
I appreciate any reference. Thanks in advance for your comments!

Answer & Explanation

Aaron Everett

Aaron Everett

Beginner2022-06-23Added 18 answers

You can just simply take another derivative to get the equation
x F + x F y 0 · y + y 1 F · y + + y n F · y ( n + 1 ) = 0
which can be transformed into an explicit ODE if y n F is invertible.
In another way, this same condition says that if y n F is invertible and continuous, then by the implicit function theorem the original implicit equation has an explicit solution
y ( n ) = g ( x , y , y , , y ( n 1 ) ) .
Usually, if y is scalar, the points where y n F = 0 form a surface and thus the set of regular points is dense, so the ability to resolve into an explicit equation is a stable property.
If y is a vector and F a system of equations of equal dimension, then the rank of y n F is a stable property. One can only resolve it into an explicit ODE if the rank is full. All other cases, excluding some more exotic degeneracies, lead to differential-algebraic equations, DAE.
An ODE has a full vector field on the state space. A DAE only defines direction vectors on a part of the state space, and then usually multiple directions per point. It becomes non-trivial to select directions so that integral curves, i.e., solutions to all defining equations, result.

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