Calculus 4

Can we say that Differential Equations is Calculus 4 in disguise?

 

The definition of the derivative involves the operation of taking a limit, and sometimes the limit problem is a derivative in disguise.

 

It is not possible to solve differential equations without knowing integration. I think this is why Differential Equations is Calculus lV in disguise.

 

The most basic differential equations are the ones which you can just integrate to get the answer.

 

I will find out in the future, Lord willing.

 

A typical Calculus textbook has, in the back so would be covered in Calculus III, has a section on separable first order differential equations, but I would NOT consider differential equations as part of Calculus itself since it covers many things that may use Calculus but are not part of Calculus itself. I would hope that, in addition to Calculus, Linear Algebra was a prerequisite for differential equations because the set of all solutions to a "homogeneous linear differential equation" forms a vector space.

 

Conceptually, calculus can include a lot more than can be discussed in two or three courses. Those introductory courses are general in nature and cover limits, derivatives, integrals, and series as well as examples and theorems relating to those concepts.
Many calculus courses do discuss a few of differential equations such as y′=ky and y′′=−y, but they don’t have time to do much. The concept is introduced, some applications, and the technique of separation of variables. That’s really no more than the first week in a course on differential equations.

 

From this I gather that differential equations is not easy at all. I know that knowledge of integration is needed to succeed in an ordinary differential equations course.

Question:

What is the basic difference between ordinary differential equations and partial differential equations?

 

An Ordinary Differential Equation is a differential equation that depends on only one independent variable. A Partial Differential Equation is differential equation in which the dependent variable depends on two or more independent variables.

 

In order to learn differential equations you certainly need to know Calculus. But the main topic of an introductory Differential Equations is "linear differential equations with constant coefficients" which are relatively easy.

For example, to solve the differential equation y''- 5y'+ 6y= 0 I would think that, for those different derivatives of y to cancel, they must be the same "type" of function- and e^x has the property that its first and second derivatives are also e^x. So I would TRY y= e^{ax}. Then y'= ae^{ax} and y''= a^2 e^{ax}. The equation becomes a^2 e^{ax}- 5ae^{ax}+ 6e^{ax}= (a^2- 5a+ 6)e^{ax}= 0. Since e^{ax} is never 0, we must have a^2- 5a+ 6= (a- 3)(a- 2)= 0 (the "characteristic equation" for this differential equation). Of course, a= 2 and a= 3 are the roots of that equation (the "characteristic roots"). So both e^{2x} and e^{3x} are solutions to the differential equation.

Now, since we have learned that the set of all solutions to an nth degree linear homogeneous differential equation form an "n dimensional linear vector space" and e^{2x} and e^{3x} are "independent functions" (one is not a constant multiple of the other), we know that ANY solution to this differential equation can be written y= Ae^{2x}+ Be^{3x} for some constants, A and B.

 

Thank you everyone for your input.