Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am taking an online course and we are currently learning Integration and this is my first time experiencing intergration, though I have some knowledge of it. I am having some difficulty understanding what a differential equation actually is. The professor defines it as

A differential equation is an algebraic equation on $x = x(t)$ and its derivatives

Then he gives the following example

$$ \frac{dx}{dt} = f(t) $$

and then the following solution

$$x(t) = \int f(t) dt$$

So does this mean that when you solve a differential equation you finding the relationship between one variable and the function of another e.g. $t$ and $x$ before $f$ was for all inputs of $t$ but now $f$ is replaced by the function of $x$. Sorry if it is difficult to read, I'm having trouble explaining my thoughts.

Any help would be appreciated


share|cite|improve this question
:see here – Maisam Hedyelloo Jun 28 '13 at 18:04

The idea of a differential equation a priori has nothing to do with integration. In the first place a differential equation is a way to condense your insight about some physical process evolving in time into a mathematical formula. Such a formula describes how the involved positions, velocities, exterior forces, etc., depend on each other at each moment of time. This means that we have a "constituent equation" of the form $$F\bigl(x(t),\dot x(t),\ddot x(t),t\bigr)=0\qquad \forall t\tag{1}$$ involving the function $t\mapsto x(t)$ and its derivatives as unknowns. "Integration" of this differential equation means that in the end we have an explicit formula for the position $x$ of some particle in function of time $t$.

An example: Assume you throw a ball vertically upwards with initial velocity $v_0$. Physical intuition tells us that (leaving friction aside) the only force acting on the ball is gravity, which exerts a constant downwards acceleration $-g$. Therefore the height $y(t)$ of the ball satisfies the differential equation $$\ddot y(t)=-g\ ,\tag{2}$$ a very simple instance of an equation of type $(1)$. "Integration" of this equation means finding a function $t\mapsto y(t)$ that satisfies $(2)$ and in addition the initial conditions $y(0)=0$, $\>\dot y(0)=v_0$. There is a unique solution, namely $t\mapsto y(t)=v_0 t-{g\over2} t^2$.

Of course an equation of the form $\dot y(t)=f(t)$ is a differential equation in the sense $(1)$, but an extremely uninteresting one. It is obvious that its solutions are the primitives of $f$.

share|cite|improve this answer
Can you clarify how can you came to $t\mapsto y(t)=v_0 t-{g\over2} t^2$? – Jeel Shah Jun 29 '13 at 18:19
@gekkostate: Using the standard technique. You'll hear about it. – Christian Blatter Jun 29 '13 at 19:26

The notion of a differential equation can be somewhat strange at first, so don't feel discouraged by not getting it instantly.

You know how in algebra you have equations like $2/x = 4$ and you want to know what (if any) numbers make this a true statement? A differential equation is like that, but instead of trying to find a number you are trying to find a function that fits.

In the example you were given you know that the first derivative of the function you are searching is equal to $f(t)$, so what do you do to get the function you actually want? Use the Fundamental Theorem of Calculus (provided $f(t)$ can be integrated), because this way you get rid of the derivative on the left hand side - and have your solution on the right hand side, the anti-derivative of $f(t)$.

share|cite|improve this answer

The important thing to note is that the unknown in a differential equation is the function, not some particular value of the variable. You're defining a relationship that some collection of functions should satisfy and then finding out what those functions are.


is saying "which functions have f(t) as their derivative?" these will be solutions to this differential equation. This is equivalent to just integrating $f(t)$, but differential equations are more general than this. Another differential equation might have the form


which is saying: "for which functions is differentiation the same as multiplying by $-1$?" Solving a differential equation is about finding the functions that satisfy the specified relationship.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.