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Since $\frac{dy}{dx}$ is not a ratio but a limit, the differential equation $\frac{d^2y}{dx^2} -3 \frac{dy}{dx} + 2y = 0$ can be interpreted as a function $y$ whose derivatives first order and second order satisfy the equation and not like differentials $dx, dy$ satisfy the equation.

But how to interpret a differential equation of the form $(z+x)dx + (x+z)dy + (x+y)dz = 0$ and also is there any rigorous way of expressing the equation above. Since a derivative must be expressed as $\frac{dy}{dx} = \lim_{h\to 0} \frac{f(x+h)- f(x)}{h}$?

Thanks in advance.

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  • $\begingroup$ I wonder... did you mean for the first term to be $(z+y)\,dy$? $\endgroup$ – amd Aug 30 '16 at 18:17
  • $\begingroup$ @amd yes.. I have made a mistake $\endgroup$ – nrynn Aug 31 '16 at 13:21
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Think of it in finite terms. Swap the differential $d$ for a finite diference $\Delta$:

$$ (z+x) \Delta x + (x+z) \Delta y + (x+y)\Delta z = 0 $$

This can be written in vectorial form as

$$ \begin{pmatrix} z+x \\ x+z \\ x+y \end{pmatrix} \cdot \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z\end{pmatrix} = 0 $$

What is it saying then?

Suppose $x = x(t)$, $y = y(t)$ and $z = z(t)$, i.e., the triplet $(x,y,x)$ defines a curve in space; then,

\begin{align} \Delta x &= x(t + \Delta t) - x(t),\\ \Delta y &= y(t + \Delta t) - y(t),\\ \Delta z &= z(t + \Delta t) - z(t),\\ \end{align}

So, the vectorial equation can be written as

$$ \begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} \frac{x(t + \Delta t) - x(t)}{\Delta t} \\ \frac{y(t + \Delta t) - y(t)}{\Delta t} \\ \frac{z(t + \Delta t) - z(t)}{\Delta t}\end{pmatrix} \Delta t = 0 $$ If the change in $t$ is infinitesimal, then $$ \begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix} d t = 0 $$ And, of course, $dt$ is as small as you like, but not zero (remember your epsilons and deltas), so

$$ \begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix} = 0 $$

Now, what on Earth is the vector on the right?

$\color{blue}{\text{The tangent vector to the curve!}}$

So, your differential equation is telling:

You need to find a curve in space such that its tangent vector is orthogonal to the vector defined by $$\mathbf{v}(t) = [z(t) + x(t)]\mathbf{i} + [x(t) + z(t)]\mathbf{j} + [x(t) + y(t)]\mathbf{k}$$ for all values of the parameter $t$.

If $\mathbf{v}(t)$ is orthogonal to $\mathbf{T}$, then it inhabits in the plane defined by $\mathcal{P} := \{\mathbf{v} \in \mathbb{R}^3\,|\, \mathbf{v}\cdot \mathbf{T} = 0\}$. In other words, $$ \mathbf{N}(t) = \mathbf{T}'(t) = \lambda \mathbf{v}(t) $$ where $\mathbf{N} \in \mathcal{P}$ and $\lambda$ is a constant.

Of course, as stated, the solution to the problem isn't unique.

But, why?

Well, suppose you have a function $u = u(x,y,z)$ and, as stated, $x = x(t)$, $y = y(t)$ and $z = z(t)$. Then, $$ \frac{d u}{d t} = u_x x'(t) + u_y y'(t) + u_z z'(t) = 0. $$

Another reading to the DE is:

You are looking for a function $u(x,y,z)$ such that its gradient is collinear to $\mathbf{v}$: $$\mu\nabla u = [z + x]\mathbf{i} + [x + z]\mathbf{j} + [x + y]\mathbf{k}$$ where $\mu$ is a constant.

So, there is an infinte number of functions that satisfy this property. Even if you focus only on the ones that pass trough a curve such that there is uniqueness for $u$, an infinite number of this curves can be constructed, ensuring lack of uniqueness for the orignial DE (even with initial conditions for $x$, $y$ and $z$ at some $t_0$).

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  • $\begingroup$ thanks a lot. So every differential equation can be rigorously written in terms of matrices and all the differentials in the equations are the derivatives with respect to the parameter. Am I correct? $\endgroup$ – nrynn Aug 31 '16 at 13:29
  • $\begingroup$ @Narayanan16 I think so, although there might be some ill examples somewhere. Also, you might need more than one parameter, but, at a reasonable level, your statement seems correct. $\endgroup$ – Pragabhava Aug 31 '16 at 15:41

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