Implicit differentiation vs differential equations?

1. Oct 16, 2012

James2

Hello, I have recently started a little implicit differentiation and I have seen DEs before but I know that I still need to work on my differentiation and integration a little more before I am ready to tackle those. Anyway, I wish to ask, what distinguishes implicit differentiation from a differential equation? In implicit differentiation you are solving an equation with a derivative in it so does that make it a differential equation even if it is comparably easier to do?

2. Oct 17, 2012

Mute

In implicit differentiation you are taking a known expression, say of two variables x and y, that cannot be expressed as a function y = f(x), and you are differentiating it with respect to one of the variables, say x, in order to solve for the derivative of the other variable, dy/dx.

For a simple example, an equation of a circle is $x^2 + y^2 = 1$. This cannot be put in the form y = f(x) for a single f(x) that describes the entire curve. If we differentiate this with respect to x, we find

$$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}.$$

Note that we started with a specific expression that related x and y and derived dy/dx.

In a differential equation, you are given an expression in terms of the dy/dx, y and x and asked to find the function y = f(x) (or even an implicit relation between the variables) which satisfies this equation.

So, for example, say you were just given the equation dy/dx = -x/y. You don't yet know anything beyond this equation. When you solve this equation, you find that $y^2 = -x^2 + C$, for some constant C. (You may be given some initial data which enables you to pin down this constant, e.g., y = 1 when x = 0 would specify C = 1).

Again, the difference here was that we had an equation for dy/dx given in terms of x and y, and we had to solve for the relationship between y and x that satisfies that differential equation.

In the first case, we had the relation between x and y, and we wanted to compute the derivative dy/dx.