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444 DIFFERENTIAL EQUATIONS<br />

Now try the following exercise.<br />

Exercise 177 Further problems on families<br />

of curves<br />

1. Sketch a family of curves represented by each<br />

of the following <strong>differential</strong> <strong>equation</strong>s:<br />

(a) dy dy dy<br />

= 6 (b) = 3x (c)<br />

dx dx dx = x + 2<br />

2. Sketch the family of curves given by the <strong>equation</strong><br />

dy = 2x + 3 and determine the <strong>equation</strong><br />

dx<br />

of one of these curves which passes through<br />

the point (1, 3). [y = x 2 + 3x − 1]<br />

46.2 Differential <strong>equation</strong>s<br />

A <strong>differential</strong> <strong>equation</strong> is one that contains <strong>differential</strong><br />

coefficients.<br />

Examples include<br />

(i) dy<br />

dx = 7x and (ii) d2 y<br />

dx 2 + 5 dy<br />

dx + 2y = 0<br />

Differential <strong>equation</strong>s are classified according to the<br />

highest derivative which occurs in them. Thus example<br />

(i) above is a first order <strong>differential</strong> <strong>equation</strong>,<br />

and example (ii) is a second order <strong>differential</strong><br />

<strong>equation</strong>.<br />

The degree of a <strong>differential</strong> <strong>equation</strong> is that of the<br />

highest power of the highest <strong>differential</strong> which the<br />

<strong>equation</strong> contains after simplification.<br />

( d 2 ) 3 ( )<br />

x dx 5<br />

Thus<br />

dt 2 + 2 = 7 is a second order<br />

dt<br />

<strong>differential</strong> <strong>equation</strong> of degree three.<br />

Starting with a <strong>differential</strong> <strong>equation</strong> it is possible,<br />

by integration and by being given sufficient data to<br />

determine unknown constants, to obtain the original<br />

function. This process is called ‘solving the<br />

<strong>differential</strong> <strong>equation</strong>’. A solution to a <strong>differential</strong><br />

<strong>equation</strong> which contains one or more arbitrary constants<br />

of integration is called the general solution<br />

of the <strong>differential</strong> <strong>equation</strong>.<br />

When additional information is given so that constants<br />

may be calculated the particular solution of<br />

the <strong>differential</strong> <strong>equation</strong> is obtained. The additional<br />

information is called boundary conditions. Itwas<br />

shown in Section 46.1 that y = 3x + c is the general<br />

solution of the <strong>differential</strong> <strong>equation</strong> dy<br />

dx = 3.<br />

Given the boundary conditions x = 1 and y = 2,<br />

produces the particular solution of y = 3x − 1.<br />

Equations which can be written in the form<br />

dy dy<br />

dy<br />

= f (x), = f (y) and = f (x) · f (y)<br />

dx dx dx<br />

can all be solved by integration. In each case it is<br />

possible to separate the y’s to one side of the <strong>equation</strong><br />

and the x’s to the other. Solving such <strong>equation</strong>s<br />

is therefore known as solution by separation of<br />

variables.<br />

46.3 The solution of <strong>equation</strong>s of the<br />

form dy<br />

dx = f (x)<br />

A <strong>differential</strong> <strong>equation</strong> of the form dy<br />

dx<br />

solved by direct integration,<br />

∫<br />

i.e. y =<br />

f (x) dx<br />

= f (x) is<br />

Problem 2. Determine the general solution of<br />

x dy<br />

dx = 2 − 4x3<br />

Rearranging x dy<br />

dx = 2 − 4x3 gives:<br />

dy<br />

dx = 2 − 4x3 = 2 x x − 4x3 = 2 x x − 4x2<br />

Integrating both sides gives:<br />

∫ ( ) 2<br />

y =<br />

x − 4x2 dx<br />

i.e. y = 2lnx − 4 3 x3 + c,<br />

which is the general solution.<br />

Problem 3. Find the particular solution of the<br />

<strong>differential</strong> <strong>equation</strong> 5 dy + 2x = 3, given the<br />

dx<br />

boundary conditions y = 1 2 when x = 2.<br />

5

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