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446 DIFFERENTIAL EQUATIONS<br />

Find the <strong>equation</strong> of the curve if it passes<br />

through the point ( 1, 1 )<br />

3 .<br />

[<br />

y = 3 ]<br />

2 x2 − x3<br />

6 − 1<br />

7. The acceleration, a, of a body is equal to its<br />

rate of change of velocity, dv . Find an <strong>equation</strong><br />

for v in terms of t, given that when t = 0,<br />

dt<br />

velocity v = u.<br />

[v = u + at]<br />

8. An object is thrown vertically upwards with<br />

an initial velocity, u, of 20 m/s. The motion<br />

of the object follows the <strong>differential</strong> <strong>equation</strong><br />

ds<br />

= u−gt, where s is the height of the object<br />

dt<br />

in metres at time t seconds and g = 9.8m/s 2 .<br />

Determine the height of the object after 3<br />

seconds if s = 0 when t = 0. [15.9 m]<br />

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

form dy<br />

dx = f ( y)<br />

A <strong>differential</strong> <strong>equation</strong> of the form dy = f (y) is<br />

dx<br />

initially rearranged to give dx = dy and then the<br />

f (y)<br />

solution is obtained by direct integration,<br />

i.e.<br />

∫<br />

∫<br />

dx =<br />

dy<br />

f ( y)<br />

Problem 6. Find the general solution of<br />

dy<br />

dx = 3 + 2y.<br />

Rearranging dy = 3 + 2y gives:<br />

dx<br />

dx =<br />

dy<br />

3 + 2y<br />

Integrating both sides gives:<br />

∫<br />

∫<br />

dx =<br />

dy<br />

3 + 2y<br />

Thus, by using the substitution u = (3 + 2y) — see<br />

Chapter 39,<br />

x =<br />

2 1 ln (3 + 2y) + c (1)<br />

It is possible to give the general solution of a <strong>differential</strong><br />

<strong>equation</strong> in a different form. For example, if<br />

c = ln k, where k is a constant, then:<br />

x =<br />

2 1 ln(3 + 2y) + ln k,<br />

i.e. x = ln(3 + 2y) 2 1 + ln k<br />

or x = ln [k √ (3 + 2y)] (2)<br />

by the laws of logarithms, from which,<br />

e x = k √ (3 + 2y) (3)<br />

Equations (1), (2) and (3) are all acceptable general<br />

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

dy<br />

dx = 3 + 2y<br />

Problem 7. Determine the particular solution<br />

of (y 2 − 1) dy = 3y given that y = 1 when<br />

dx<br />

x = 2 1 6<br />

Rearranging gives:<br />

( y 2 − 1<br />

dx =<br />

3y<br />

Integrating gives:<br />

∫<br />

dx =<br />

) ( y<br />

dy =<br />

3 − 1 )<br />

dy<br />

3y<br />

∫ ( y<br />

3 − 1 3y<br />

)<br />

dy<br />

i.e. x = y2<br />

6 − 1 ln y + c,<br />

3<br />

which is the general solution.<br />

When y = 1, x = 2 1 6 , thus 2 1 6 = 1 6 − 1 3<br />

ln 1 + c, from<br />

which, c = 2.<br />

Hence the particular solution is:<br />

x = y2<br />

6 − 1 3 ln y + 2<br />

Problem 8. (a) The variation of resistance,<br />

R ohms, of an aluminium conductor with<br />

temperature θ ◦ C is given by dR = αR, where<br />

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