[forward][up]

7.1 introduction

A differential equation (DE) is any equation with a differential in it.

Examples: (a) {dy\over dx} = 1, (b) {dy\over dx} = {x}^{2}, (c) {dy\over dx} = y, (d) x{dy\over dx} + 2y =\mathop{ cos}\nolimits x, (e) {{d}^{2}y\over { dx}^{2}} + 2{dy\over dx} + 3y = 0.

Differential equations occur in many models of real-world situations. One particular examples when we consider rates of change, e.g., (f) the concentration in C a chemical reaction {dC\over dt} = a − kC, (g) (explosive) population growth {dN\over dt} = αN, (h) simple harmonic motion {{d}^{2}x\over { dt}^{2}} = −{ω}^{2}x, (i) motion under the influence of the earth gravitional field, m{{d}^{2}y\over { dt}^{2}} = −mg.

We would really like to classify such equations by their order.

The order of a DE is the highest derivative contained in it.
Thus (a), (b), (c), (d), (f), (g) are first order, and (c), (h), (i) are second order. In this course we only consider first order DEs.

Solving DEs is sometimes called integrating them, since for the simplest types this is exactly what we do. Just as for integration we draw up a list of standard types that we know how to do.

Most solutions of DEs contain constants. These are just like constants of integration, and arise from the fact that the derivatives of these constants is 0. We always get as many arbitrary constants as the order of the equation. The general solution will include these arbitrary constants. If we have extra information apart from the DE itself we can find the arbitrary constants. This extra information is sometimes called “initial conditions” or “boundary conditions”. Once the arbitrary constants are known we have the actual solution.

We shall study several types of DEs to facilitate solution, but let’s first look at two simple examples.

Example 7.1: 

Solve the DE

{dy\over dx} = {x}^{2} − 2,

given that y = 1 for x = 0.

Solution: 

Integrate both sides of the equation,

y =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dy\over dx}dx =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits ({x}^{2} − 2)dx = {1\over 3}{x}^{3} − 2x + k.

At x = 0, y = 1, which implies k = 1. Thus

\class{boxed}{y(x) = {{x}^{3}\over 3} − 2x + 1. }

Example 7.2: 

Find the general solution of \mathop{cos}\nolimits x\kern 1.66702pt dy∕dx + 2\mathop{sin}\nolimits x = 0.

Solution: 

Rearrange as

\begin{eqnarray*} {dy\over dx}& =& −2 {\mathop{sin}\nolimits x\over \mathop{cos}\nolimits x} %& \\ & =& −2\mathop{tan}\nolimits x\quad .%& \\ \end{eqnarray*}

Integrate

\class{boxed}{y = −2\mathop{ln}\nolimits (\mathop{sec}\nolimits x) + k. }

[forward][up]