# Understanding differential equations

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

Thanks!

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: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 $f$.

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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 $f(t)$.

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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.

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

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

$$\frac{dx}{dt}=-x$$

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.

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