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So the first step i want to learn before mastering differential equations is the classification of differential equations. In general, a differential equation is said to be an equation involving an unknown function (dependent variable ) and its derivatives with respect to one or more independent variables.

Is $$d^2x/dt^2 + dx/dt = d^2y/dt^2 + y$$ a differential equation?

Does it satisfy the definition given by wikipedia ( A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders ) ?

I don't think so because the definition mentions ONE uknown function, the case i just showed has 2 unknown functions and its derivatives.

Shouldn't the definition be :

A differential equation is a mathematical equation for ONE OR MORE unknown functions of one or several variables that relates the values of the FUNCTIONS THEMSELVES and its derivatives of various orders with respect to one or more independent variables

?

Clarify my doubt please.

Pedro
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user1843665
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  • So why doesn't it fit the definition by any book and by wikipedia ? – user1843665 Mar 09 '13 at 01:58
  • The definition should be : A differential equation is a mathematical equation for ONE OR MORE unknown functions of one or several variables that relates the values of the FUNCTIONS THEMSELVES and its derivatives of various orders with respect to one or more independent variables – user1843665 Mar 09 '13 at 01:59
  • In general, the pair of functions $x(t)$ and $y(t)$ can be seen as a single function $t\to (x(t),y(t))$. The problem with Wikipedia's definition is that it doesn't say the range can be multi-dimensional, only the variables. – Thomas Andrews Mar 09 '13 at 02:01
  • Yes, i know the differential equation i mentioned is the solution to this system : [d²x/dt²+dx/dt = 0] and [ - ( d2y/dt2+y ) = 0 ]

    But what bothers me is the differential equation definition given by almost every book and even by wikipedia

     – user1843665 Mar 09 '13 at 02:02
  • It is one function, but it is a function from $\mathbb R$ to $\mathbb R^2$, as I suggested above in the previous comment. In general, several functions of the same domain can be seen as one function which returns an ordered list of values. – Thomas Andrews Mar 09 '13 at 02:04
  • It is definitely not a dumb question, pointing as it does to a flaw, or at least lack of clarity, in the Wikipedia definition. – André Nicolas Mar 09 '13 at 02:05
  • I agree with Andre, it is confusing, although I don't think Wikipedia is wrong so much as it makes an advanced assumption. – Thomas Andrews Mar 09 '13 at 02:09
  • 7 Definitions & a Related Question – sponsoredwalk Mar 09 '13 at 02:23
  • What about the n-tuple of functions ( x^n (t), x^(n-1) (t) , ... , x(t) , y^n (t), y^(n-1) (t) , ..., y(t) ) is it just only one function as well ? Can we say a differential equation is the same as equating a function to zero ? – user1843665 Mar 09 '13 at 02:23
  • And by the way, assuming you have a function made by many other functions and saying that the differential equation is given by this function and by the derivatives of this function would imply we would only have derivatives of (x(t),y(t)) which is very different from the derivatives of x(t) and y(t) particularly – user1843665 Mar 09 '13 at 02:32

2 Answers2

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Yes, it is a differential equation, never mind what Wikipedia may say.

Typically, one meets differential equations like that one when studying systems of differential equations. Just like typically you meet equations like $3x+4y+5z=17$ when studying systems of equations.

André Nicolas
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Thomas, you are saying 4x(t) + 3y(t), for example, is only one function, right ? So, by the same token, we can make the entire differential equation only one function as well.The differential equation would be equating this master function to zero :

t -> ( x^n (t), x^(n-1) (t) , ... , x(t) , y^n (t), y^(n-1) (t) , ..., y(t) )

user1843665
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  • No, I'm saying $(x(t),y(t))$ is only one function. A function is a map from one set to another set. While we have two functions $x:\mathbb R\to \mathbb R$ and $y:\mathbb R\to \mathbb R$, we can also see them as one function $(x,y):\mathbb R\to \mathbb R^2$. – Thomas Andrews Mar 09 '13 at 02:40
  • why can't se see x: R→R,y:R→R and dx/dt: R→R as only one function that maps R→R3 ? – user1843665 Mar 09 '13 at 03:24
  • By the same token we can consider the entire nth order differential equation as one function that maps R→Rw – user1843665 Mar 09 '13 at 03:27
  • OH, yes, but you wrote: "you are saying $4x(t)+3y(t)$ for example, is only one function, right?" Which was not what I was saying at all. – Thomas Andrews Mar 09 '13 at 03:30
  • Ket $z(t)=(x(t),y(t))$. Then $z^{(n)}(t)=(x^{(n)}(t),y^{(n)}(t))$ and we can write the entire formula as a function $h(z,z',z'')=0$. It uses the natural projections $\pi_1,\pi_2:\mathbb R^2\to\mathbb R$, where $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. – Thomas Andrews Mar 09 '13 at 03:33
  • So if we define $$h(u,v,w) = \pi_1(w) + \pi_1(v) - \pi_2(u) -\pi_2(v)$$ then $h(z,z',z'') = 0$ is your differential equation. – Thomas Andrews Mar 09 '13 at 03:35
  • So the supposed-advanced definition by wikipedia is wrong because there is not a single derivative of h in the differential equation.It says a differential equation has one function ( h ) and derivatives of this function but the differential equation is h = 0, theres no h',h'' etc – user1843665 Mar 09 '13 at 05:21
  • Sigh, the single function is $z$. – Thomas Andrews Mar 09 '13 at 05:39
  • Perhaps I shouldn't have added $h$, and instead just written the DE as $\pi_1(z'') + \pi_1(z') -\pi_2(z)-\pi_2(z')=0$. The point is, we are trying solve an equation in one function, $z$. There are other functions involved, like $\pi_1$ and $\pi_2$, but they are constants - they are knowns. The $z$ is the unknown, and there is only one unknown function in the equation. (By that point, $h$ above is a "known" function - we have already defined it.) – Thomas Andrews Mar 09 '13 at 05:45