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An Introduction to Differential Equations

First-Order Homogeneous Linear Diff Eqs Tutorial Index
First-Order Nonhomogeneous Linear Diff Eqs Maple Index
A System with Air Resistance How to...
Problem Set Index

Let's start with a definition of a differential equation.  A differential equation is an equation that defines a relationship between a function and one or more derivatives of that function.  Let y be some function of the independent variable t.  Then following are some differential equations relating y to one or more of its derivatives.

The equation

d/dt y(t)=t^2 y(t)

states that the first derivative of the function y equals the product of t^2and the function y itself.  An additional, implicit statement in this differential equation is that the stated relationship holds only for all t for which both the function and its first derivative are defined.  Some other differential equations:

d^2/dt^2 y(t)=-y(t)

5 d^2/dt^2 y(t)+3/2 d/dt y(t)+1/2 y(t)=sin(t)

d/dt y(t)=y(t)^2

The typical real world situation that involves modeling with differential equations is one in which a quantity that changes in time, y(t), is related to the rate at which it changes, d/dt y(t), and/or the acceleration of the change, d^2/dt^2 y(t).  Usually these relationships can be stated in equations such as those above without knowing the exact function y(t) that satisfies the relationship.  To solve a differential equation means to find a continuous function of the independent variable that, along with its derivatives, satisfies the equation.  For example, the natural log function, ln, is a solution to the following differential equation.

d/dt y(t)=1/e^y(t)

The symbol e is the base of the natural log.  If you do not see why ln satisfies this equation, take a moment to figure it out (substitute ln(t) into the equation for y(t) and then simplify it).

In differential equations the unknowns of concern are no longer unknown variables, but unknown functions that stand in certain mathematical relationships with their derivatives, other functions, constants, and regular variables.   Just as in the more familiar setting of equations describing the relationships between variables like x, y, and z, these relationships can be stated in equations.   And also as with equations relating variables, techniques exist for solving them.   And to continue the analogy yet some more, just as the equations with which you are familiar are classified into types according to their forms and solution techniques, so to with differential equations.

To continue the analogy yet even a little more, just as was the case many years ago when you first started solving equations of variables, you will start solving differential equations by learning to solve the linear types (actually, for some of you, depending on your calculus instructor, you may have already seen another type of differential equation, separable).  The reason is that just as in the case of regular linear equations, linear differential equations are easier to solve than most other types of differential equations (if you are interested in the learning some of the many different types of differential equations, check out the help pages linked to from ?odeadvisor).  However, linear differential equations are more difficult to solve than simple linear equations, and this, of course, is why you're learning about them now rather than many years ago.  But, you should have no difficulty understanding the solution technique, because no new mathematics are involved; each step uses only mathematical ideas with which you are already familiar.

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Before moving on to the details of linear differential equations and how to solve them, note that you already know how to solve some differential equations.  For example, you can certainly solve the following differential equation.

d/dt y(t)=t

Simply integrate both sides of it.

y(t)=1/2 t^2+C

With some simple algebraic manipulations, both of the following differential equations can be transformed into the same form as the one above, and then solved in the exact same way -- integrate both sides of the equations.

d/dt y(t)+2 t-5=6

has solution

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And

e^t+ d/dt y(t)-a^2=0

has solution

y(t)=-e^t+a^2+C

The key idea here is that you know how to solve differential equations that can be placed in the form

d/dt some function of t = all other parts of the equation

Of course, in all these cases only the derivative of the unknown function appears in the equation, but the same principle will apply in the solution strategies of linear differential equations in which both the unknown function and its derivative appear.  Keep this in mind.  Now read First-Order Homogeneous Linear Differential Equations.

 

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