Transcript First Order Linear Differential Equations

First Order Linear Differential
Equations
Chapter 4 FP2 EdExcel
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Splitting the Variables
This is a method used in C4 of the Core
Mathematics A level and then in Chapter 4
FP2.
dy
When:
 f ( x) g ( y )
dx
We can split the variable in order to find a
final equation in the form : y=f(x)
We do this by Dividing and Multiplying to get
g(y) on one side and f(x) on the other side,
and splitting dy/dx up to the appropriate
sides like so:
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 g ( y) dy   f ( x)dx
We can now integrate to find y in the form
f(x). Do not forget the arbitrary constant
when doing so, and if given values for x and
y, use them to find the value of this constant.
•
Worked example (Elmwood C4, paper A)
dy
 y 2 ( 4 x 5  1)
dx
1
4 x5  1
dy 
dx
y2
x2
x2
1
4 x5
1
dy


 y2
 x 2 x 2 dx
1

  4 x 3  x  2 dx
y
1
1

 x4   C
y
x
1
x5  1


C
y
x
x
y 5
C
x 1
2
dy/dx +p(x)y=q(x)
Multiplying Factors
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The problem in such a differential equation is that
one can not simply divide and multiply to split the
variables. We therefore need to manipulate the
expression to allow calculus to be used to solve the
problem.
In order to do this we need to think back to C3
calculus and remember the product rule.
du
dv d
v
u
 (uv)
dx
dx dx
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We need to manipulate the equation into this form,
and we can do this by using a multiplying factor,
M(x).
By doing this, we have allowed ourselves to
manipulate the equation into the form of the product
rule, because we can let M’(x)=M(x)q(x), and thus
find M(x).
M ( x)  M ( x) p ( x)
M ( x)
 p( x)
M ( x)
M ( x)
 M ( x) dx   p( x)dx
ln M ( x) 
M ( x) dy
dx  M ( x ) p ( x ) y  M ( x ) q ( x )
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We know that:
 p( x)dx
p ( x ) dx

M ( x)  e
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This gives us a generalised form of our Multiplying
factor M(x). We can ignore arbitrary constants in
finding M(x) as we only need a solution, not a
particular one, and they would only cancel in the
equation anyway.
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Using the Multiplying factors
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We know that:
M ( x)
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dy
 M ( x) p( x) y  M ( x)q ( x)
dx
We also know that:
p ( x ) dx

M ( x)  e
•
We now get the form:
M ( x)
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dy
 M ( x) y  M ( x)q ( x)
dx
The LHS takes the form of the product rule
and can be changed to:
d
 y.M ( x)
dx
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So we get:
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Which can be solved using calculus.
d
y.M ( x)  M ( x)q( x)
dx
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Worked example (EdExcel FP1 Jan ’08):
dy
 3y  x
dx
dy
M ( x)  M ( x)3 y  M ( x) x
dx
p ( x ) dx
M ( x)  e 
M ( x )  e 3 x
dy
e 3 x
 3e 3 x y  xe3 x
dx
d 3 x
(e . y )  xe3 x
dx
xe3 x e 3 x
3 x
e .y  

 C..........by. parts
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x 1
y     Ce3 x
3 9
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Substitutions
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One can use substitutions in order to transform
differential equations into more easily solvable
problems. These substitutions eliminate y from the
equation and create a differential equation in terms
of x and z (where z is the substitution).
dy x 2  3 y 2

dx
2 xy
y
let _ z   y  xz
x
dy
dz
x z
dx
dx
dz
x 2 (1  3 z 2 )
x z
dx
2x2 z
dz (1  3 z 2 )
x 
z
dx
2z
dz 1  z 2
x 
»
dx
2z
This can now be solved by
splitting the variables as shown
on page 2.
One can also use substitutions in order to eliminate
x and y and create a simple calculus problem to
solve.
dy y  x  2

dx y  x  3
let _ u  y  x
du dy

1
dx dx
du
u2
1 
dx
u 3
du u  2

1
dx u  3
du u  2  u  3

dx
u 3
du
1
» This can now be solved by

splitting the variables as shown
dx u  3
on page 2.
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