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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:

$${(y^{(1)})}^5+sen(y^{(1)})+e^{y^{(1)}} + x=0$$

I appreciate any reference. Thanks in advance for your comments!

01mgfx
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  • The conclusion is that not all ODE's are of the above form, except perhaps locally with approximations. There is nothing more to draw from it, and you will rarely be able to solve such ODEs. – Alex R. Feb 06 '16 at 01:54
  • The differential equations that we are interested in are usually of the form $y'=f(x,y)$. Now if it is not if that firm then you might rarely get and answer due to the obscure integral that you will end up getting – Sayan Chattopadhyay Feb 06 '16 at 02:44
  • What is the $sen$ function? – doraemonpaul Feb 06 '16 at 15:10

3 Answers3

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If you cannot express your derivative then you deal with an implicit differential equations They are very important and extensively studied. Depending on what is your level of preparation you can find differnt accounts how to deal with such equations. A particularly interesting and very geometric treatment of such equations is given in Arnold's Geometric methods in the theory of ODE.

Here is just one example how the solutions can look like.

Community
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Artem
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You can just simply take another derivative to get the equation $$ ∂_xF+∂_xF_{y_0}·y'+∂_{y_1}F·y''+…+∂_{y_n}F·y^{(n+1)}=0 $$ 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.

Lutz Lehmann
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For $(y')^5+sen(y')+e^{y'}+x=0$ it is just an ODE of the form $f(y')+x=0$ .

Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2:

Let $P(x,y,t)=f(t)+x$ ,

$\because\dfrac{dx}{dt}=-\dfrac{\dfrac{\partial P}{\partial t}}{\dfrac{\partial P}{\partial x}+t\dfrac{\partial P}{\partial y}}=-f'(t)$ or $\dfrac{dy}{dt}=-\dfrac{t\dfrac{\partial P}{\partial t}}{\dfrac{\partial P}{\partial x}+t\dfrac{\partial P}{\partial y}}=-tf'(t)$

This ODE can still solve easily.

doraemonpaul
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