FET 222
PART A1
FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS
INTRODUCTION
A differential equation is a relationship between an independent variable, x , a dependent variable, y , and
one or more differential coefficients of y with respect to x.
Differential equations represent dynamic relationships, i.e. quantities that change, and are thus frequently
occurring in scientific and engineering problems.
FORMATION OF DIFFERENTIAL EQUATIONS
Differential equations may be formed in practice from a consideration of the physical problems to which
they refer. Consider an example.
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Let T(t) be the temperature of a body and TA be the air temperature (assumed constant) . Then by
Newton’s law
dT
 k T  TA 
dt
SOLUTION OF DIFFERENTIAL EQUATIONS
To solve a differential equation, we have to find the function for which the equation is true. This means that
we have to manipulate the equation so as to eliminate all the differential coefficients and leave a
relationship between y and x or t
Various methods are used in solving first order differential equations depending on the form of the
equation. These include
1. By direct integration:
dy
 f x
dx
gives y   f  x  dx
2. By separating the variables: g  y 
gives
dy
 f x
dx
 g  y  dy   f  x  dx
3. Homogeneous equations: Substitute y = vx
2
gives
vx
dv
 g v 
dx
4. Exact equations: M  x, y  dx  N  x, y   0
solution u  x, y   0
5. Linear equations:
dy
 P xy  Qx
dx
Gives y 
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Q.IFdx
IF 
where IF  e 
6. Bernoulli's equation:
Divide by y n
Pdx
dy
 P xy  Qxyn
dx
: then put z  y 1n
Reduces to form 5 above.
METHOD 1
DIRECT INTEGRATION
If the equation can be arranged in the form,
dy
 f  x  , then the equation can be solved by simple
dx
integration as follows
 dy   f  x dx
This gives
y   f  x dx
Please see FET 222 A1 worked examples 1, 2 and 3
METHOD 2
SEPARABLE EQUATIONS
If the given equation is of the form-
dy
 f  x, y  the variable y on the right-hand side, prevents solving by
dx
direct integration. We therefore have to devise some other method of solution.
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Let us consider equations of the form
dy
 f xg y 
dx
and of the form
dy f  x 

dx g  y 
,i.e. equations in which the right-hand side can be expressed as products or quotients of functions of x or of
y.
The method depends on our being able to express the given equation in the form
g y 
dy
 f x
dx
If this can be done, the rest is then easy, for then we have
dy
 g  y  dx dx   f  x  dx .
Therefore
 g  y  dy   f  x  dx
We then continue by integrating each side with respect to its variable.
Please see FET 222 A1 worked examples 4, 5, 6, 7, 8, 9, 10 and 11
METHOD 3 HOMOGENEOUS EQUATIONS
Consider the equation :
It is an example of a homogeneous ODE. This is determined by the fact that the total degree in x and y for
each of the terms involved is the same (in this case, of degree 1). The key to solving every homogeneous
equation is to substitute y = vx where v is a function of x. This converts the equation into a form in which
we can solve by separating the variables.
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Please see FET 222 A1 worked examples 12, 13 and 14
METHOD 4 EXACT EQUATIONS
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6
Please see FET 222 A1 worked examples 15, 16 and 17.
METHOD 5 LINEAR EQUATIONS
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The integrating factor IF is given is giving by
IF  e 
Pdx
When we multiply both sides by the integrating factor IF the L.H.S. is converted into a complete differential
coefficient.
So : To solve a differential equation of the form
dy
 Py  Q
dx
where P and Q are constants or functions of x, multiply both sides by the integrating factor
IF  e 
Pdx
This gives
e
pdx
pdx
pdx
dy
 Pye   Qe 
dx
We now find that the L.H.S. is, in fact, the differential coefficient of
ye 
pdx
That is
e
pdx
pdx
dy
d   pdx 
 pdx
 Pye  
 ye
  Qe
dx
dx 

Now, of course, the rest is easy. Integrate both sides w.r.t. x:
ye 
pdx
  Qe 
pdx
C
Or simply
y.IF   Q.IFdx  C
Hence
y  IF 1   Q.IFdx  C 


where
IF  e 
Pdx
Note that in determining  Pdx ,we do not include a constant of integration. This omission is purely for
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convenience, for a constant of integration here would in fact give a constant factor on both sides of the
equation, which would subsequently cancel. This is one of the rare occasions when we do not write down
the constant of integration.
Please see FET 222 A1 worked examples 18, 19, 20, 21, 22, 23 and 24.
METHOD 6 BENOULLI EQUATIONS
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Please see FET 222 A1 worked examples 25, 26, 27 and 28.
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