Lovellss

2022-06-22

An ODE (Ordinary Differential Equation) of order n becomes a relation:
$F\left(x,y,{y}^{\left(1\right)},...,{y}^{\left(n\right)}\right)=0$
Then $F\left(x,y,{y}^{\left(1\right)}\right)=0$ 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:
${y}^{\left(1\right)}=f\left(x,y\right)$
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:
${\left({y}^{\left(1\right)}\right)}^{5}+sen\left({y}^{\left(1\right)}\right)+{e}^{{y}^{\left(1\right)}}+x=0$

Aaron Everett

You can just simply take another derivative to get the equation
${\partial }_{x}F+{\partial }_{x}{F}_{{y}_{0}}·{y}^{\prime }+{\partial }_{{y}_{1}}F·{y}^{″}+\dots +{\partial }_{{y}_{n}}F·{y}^{\left(n+1\right)}=0$
which can be transformed into an explicit ODE if ${\partial }_{{y}_{n}}F$ is invertible.
In another way, this same condition says that if ${\partial }_{{y}_{n}}F$ is invertible and continuous, then by the implicit function theorem the original implicit equation has an explicit solution
${y}^{\left(n\right)}=g\left(x,y,{y}^{\prime },\dots ,{y}^{\left(n-1\right)}\right).$
Usually, if y is scalar, the points where ${\partial }_{{y}_{n}}F=0$ 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 ${\partial }_{{y}_{n}}F$ 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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