What exactly is a differential equation? Pray, is $$\frac{dx}{dt}= x(x(t))$$ not a DE?  Because Arnold thinks not. In a note on page 11 of his Ordinary Differential Equations, he says, 'Differential equations are sometimes said to be equations containing unknown functions and their derivatives. This is false...'  And this is -- I remember -- exactly what we were told during our undergraduate ODE classes; I also have found this in many other places -- books, online, etc. So Arnold's comment is strange and seems like nonsense.  Who has an idea what he means, or why he's saying this?
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Here is one definition of a differential equation:  &quot;An equation containing the derivatives of one or more dependent variables, with respect to one of more independent variables, is said to be a differential equation (DE)&quot; (Zill - A First Course in Differential Equations)  Here is another:  &quot;A differential equation is a relationship between a function of time &amp; it's derivatives&quot; (Braun - Differential equations and their applications)  Here is another:  &quot;Equations in which the unknown function or the vector function appears under the sign of the derivative or the differential are called differential equations&quot; (L. Elsgolts - Differential Equations &amp; the Calculus of Variations)  Here is another:  &quot;Let $f(x)$ define a function of $x$ on an interval $I: a &lt; x &lt; b$. By an ordinary differential equation we mean an equation involving $x$, the function $f(x)$ and one of more of it's derivatives&quot; (Tenenbaum/Pollard - Ordinary Differential Equations)  Here is another:  &quot;A differential equation is an equation that relates in a nontrivial way an unknown function &amp; one or more of the derivatives or differentials of an unknown function with respect to one or more independent variables.&quot; (Ross - Differential Equations)  Here is another:  &quot;A differential equation is an equation relating some function $f$ to one or more of it's derivatives.&quot; (Krantz - Differential equations demystified)  Now, you can see that while there is just some tiny variation between them, calling $f(x)$ the function instead of $f$ or calling it a function instead of an equation but generally they all hint at the same thing. However:  &quot;Let $U$ be an open domain of n-dimensional euclidean space, &amp; let $v$ be a vector field in $U$. Then by the differential equation determined by the vector field $v$ is meant the equation $x' = v(x), x \in U$. Differential equations are sometimes said to be equations containing unknown functions and their derivatives. This is false. For example, the equations $\frac{dx}{dt} = x(x(t))$ is not a differential equation.&quot; (Arnold - Ordinary Differential Equations)  This is quite different &amp; the last comment basically says that all of the above definitions, in all of the standard textbooks, are in fact incorrect. Would anyone care to expand upon this point if it is of interest as some of you might know about Arnold's book &amp; perhaps be able to give some clearer examples than $\frac{dx}{dt} = x(x(t))$, I honestly can't even see how to make sense of $\frac{dx}{dt} = x(x(t))$. The more explicit (and with more detail) the better! A second question I would really appreciate an answer to would be - is there any other book that takes the view of differential equations that Arnold does? I can't find any elementary book that starts by defining differential equations in the way Arnol'd does &amp; then goes on to work in phase spaces etc... . Multiple references welcomed.
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Given the following equation:  $f(xy)=f(x)+f(y)$  and the fact that $y=x^{-1}$,  I've to find how this could became:  $f'(x) = f'(1)/x$,  where it is said that $f'(x)$ is the total derivative of $f$  I really miss some steps..someone could help?
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In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.  The first issue I have is that my book began with the statement that given  $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  the problem $L[y] \ = \ f$ can be re-cast in the form $L[y] \ = \ \lambda y$.  Now it could be a typo on their part but I see no justification for the way you can just do that!  More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:  Given a linear second order ode   $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  it is an exact equation if it is derivable from a differential equation of one order lower, i.e.  $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$  The equation is exact iff  $$a''(x) - b'(x) + c(x) = 0. $$  If $F$ is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.  This equation is exact iff  $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$  If you expand this out you get the Adjoint operator  $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$  If you expand $L^*$ you see that we can satisfy $L \ = \ L^*$ if $a'(x) \ = \ b(x)$ &amp; $a''(x) \ = \ b'(x)$ which then turns $L[y]$ into something of the form  $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$  Thus we seek an integratiing factor $\alpha(x)$ so that we can satisfy this &amp; the condition this will hold is that $\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}$  Then we're dealing with:  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$  But again, by what my book said they magically re-cast this problem as  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$  Then calling  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$  a Sturm-Liouville problem.  My question is, how can I make sense of everything I wrote above? How can I clean it up &amp; interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues &amp; finding full solutions - what's going on? I want to be able to look at $a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)$ &amp; know how &amp; why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness &amp; integrating factors, thanks for reading!
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In order to solve  $$y''-2y'+y=te^t$$ I did:  $$y = (at+b)e^t\implies\\ y' = ate^t+ae^t+be^t\implies \\y''= ate^t+2ae^t+be^t$$  but when I put these in the differential equation I'm getting zeros eveywhere. Does somebody know what I'm doing wrong?
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Answers to this question  assert that to find a solution to the differential equation $$\dfrac{dy}{dx} = \dfrac{y}{x}$$ we may rearrange and integrate $$\int \frac{1}{y}\ dy=\int \frac{1}{x}\ dx.$$  If we perform the integration we get $\log y=\log x+c$ or $$y=kx$$ for constants $c,k \in \mathbb{R}$.  I've seen others use methods like this before too, but I'm unsure why it works.  Question: Why is it legitimate to solve the differential equation in this way?
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I have a differential equation system  $x_1'(t) = -x_2(t)$  $x_2'(t) = -x_1(t)$  I see that I can write $\dot{x} = Ax$ where  $A = \begin{pmatrix}0 &amp; -1 \\ -1 &amp; 0\end{pmatrix}$  The complete real solution will be  $x(t) = c_1 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix} + c_2 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}$  where $c_1,c_2 \in \mathbb{R}$.  It this correct?  Now I need to find the solution such that $x_1(0) = -1$ and $x_2(0) = 1$.  How can I do this?  I think I could split my complete solution in  $x_1(t) = c_1 e^t - c_2 e^{-t}$  and  $x_2(t) = c_1 e^t + c_2 e^{-t}$  and then find $c_1$ and $c_2$ by solving $x_1(0) = -1$ and $x_2(0) = 1$ with respect to $c_1$ and $c_2$ and finally substitute these values in the complete solution $x(t)$.  Is this correct?
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I read the following statement in a book on Calculus, as part of my mathematics course:  Technically this separation of $\frac{dy}{dx}$ is not mathematically valid. However, the resulting integration leads to correct answer.  The book also contains the following:  To solve a differential equation by separation of variables:  get all the $x$ values on one side and all the $y$ values on the other side by multiplication and division. separate $\frac{dy}{dx}$ as if it were a fraction. integrate both sides.   Note: This box doesn't refer to a particular problem. It refers to a class of problems of differential equations which can be solved using the Method of Separation of Variables. My high school mathematics teacher told me that this is the most fundamental way to solve differential equations but the textbook says it is not mathematically valid. I am not able to understand why are certain methods being followed without having a mathematical proof. Or am I wrong?
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$$\frac{d}{dx} e^x=e^x$$ Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.
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im not good at formatting. May I know why it’s false   $x^2= x+x+x+\ldots(x\; \text{times})$  apply derivative on both sides  $=&gt; \frac{d}{dx}(x^2)=\frac{d}{dx}(x+x+x+\ldots(x\; \text{times}))$  $=&gt; 2x=1+1+1+\ldots(x\; \text{times})$  $=&gt; 2x=x$  $=&gt; 2=1$
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I know that $\frac{dy'(x)}{dy}=0$ (where $y'=\frac{dy(x)}{dx}$). The reason explained is that $y'$  does not depend explicitly on $y$. But intuitively, $y'$ depends on $y$, since if you vary $y$ you will modify $y'$. Why is my reasoning wrong (my reasoning sounds like it's related to functional calculus, instead of standard calculus)?  I tried to write  $\frac{dy'(x)}{dy}=\frac{d}{dy}\frac{dy}{dx}=\frac{d}{dx}\frac{dy}{dy}=\frac{d}{dx}1=0$, but this proof doesn't convince me.  I think that other way to see this is saying that if a function is of the form $f(y)$, it will not dependend on the variable $y'$.  But the same way you write $f(y)=y^2$, you could write $f(y)=\frac{d}{dx}y$, which clearly depends on $y'$. So I don't know if there are some types of operations which are restricted (for example, taking limits):  Note: The problem raises in the context of Classical Mechanics, where: $\frac{\partial L(\dot{x},t)}{\partial x}=0$.
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In school I was taught that we use $\frac{du}{dx}$ as a notation for the first derivative of a function $u(x)$. I was also told that we could use the $d$ just like any variable. After some time we were given the notation for the second derivative and it was explained as follows: $$ \frac{d(\frac{du}{dx})}{dx} = \frac{d^2 u}{dx^2} $$ What I do not get here is, if we can use the $d$ as any variable, I would get the following result: $$ \frac{d(\frac{du}{dx})}{dx} =\frac{ddu}{dx\,dx} = \frac{d^2 u}{d^2 x^2} $$ Apparently it is not the same as the notation we were given. A $d$ is missing. I have done some research on this and found some vague comments about &quot;There are reasons for that, but you do not need to know...&quot; or &quot;That is mainly a  notation issue, but you do not need to know further.&quot; So what I am asking for is: Is this really just a notation thing? If so, does this mean we can actually NOT use d like a variable? If not, where does the $d$ go? I found  related question, but it does not really answer my specific question. So I would not see it as a duplicate, but correct me if my search has not been sufficient and there indeed is a similar question out there already.
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Just had a quick question regarding the uniqueness of the general solution of first-order differential equations:  For $x'(t)=Ax(t)$ where $A$ is an $n\times n$ matrix. and initial condition $x(0)=x_0$, the unique solution is $x(t)=e^{tA}x_0$.  The proof for showing that the solution is unique involves differentiating the matrix product $e^{-tA}x(t)$.  Could someone please explain why we always multiply by $e^{-tA}$.   Many thanks, Best.
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The solution of differential equation $\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$   $\bf{My\; Try::}$ Let $\displaystyle \frac{dy}{dx} = t\;,$ Then Diferential equation convert into $t^2-xt+y=0$  So Its solution is given by $\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}$  So we get $$\frac{dy}{dx} = \frac{x\pm \sqrt{x^2-4y}}{2}$$  Now How can I solve after that, Help me  Thanks
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For an $n$th order differential equation, why are there always $n$ solutions? Why exactly $n$, not $n - 1, n+1$ or infinite many?    Addendum by LePressentiment :   This is motivated by P176 on Strang's Intro to Lin Alg, 4th Ed. An $n$th order differential equation, how many basis functions does it have? I'd guess $n$, because the diffl eqn have $n$ linearly independent solutions (predicated on Julián Aguirre's answer), all of which look to span the nullspace/solution space of the homogeneous ODE.
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What is the general solution to the differential equation:  $$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}$$  I'm a little stuck because all the techniques I know are unable to solve it.
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I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:  So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get.  Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?  And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.  Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle.   I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.  Any assistance?
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Today I read the answer to , in which the poster integrates $x^5e^x$ by making these manipulations with the differential operator $D$: $$\frac1Dx^5e^x=e^x\frac{1}{1+D}x^5=e^x(1-D+D^2+...)x^5$$  which I was amazed by but yet suspicious of. After reading a bit on differential operators, I know a few properties. For example, $D+a$ (where $a$ is constant) is a polynomial differential operator which comes from the differential equation $y'+ay = q(x)$. Also, for polynomials $f$ and $g$, $f(x)\cdot g(x)=h(x)$ implies $h(D)u=f(D) \circ [g(D)u]$ with $u$ being a function and $f(D), g(D), h(D)$ being polynomial differential operators. Since $(1+x) \cdot \frac1{1+x} = 1$, I know that if we applied the operator $(1+D)$ to some function, $\frac{1}{1+D}$ would invert it back to the original function.  However, this still doesn't help me make sense of the meaning of $\frac{1}{1+D}$. For example:   If $1+D$ comes from the differential equation $y'+ay+q$, where does $\frac{1}{1+D}$ come from? In other words, if $(1+D)y = 1\cdot y + Dy = y' + y$, how do we compute $\frac{1}{1+D}y$? (Is it even possible, and how does it relate to integration?) How do we justify the power series of the operator? How do we know there are no convergence issues?