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FirstOrder Linear Differential Equations #2
Generating New Differential Equations using a Multiplier Function
In this section we show how to obtain new firstorder differential equations by transforming the solutions of a given
firstorder DE by multiplying them by a given function
m
(
t
). One application of this theory is that the new differential
equation is easier to solve if the multiplier is chosen judiciously.
Example
: In a previous section, we saw the solutions of the 1
st
order
differential equation:
dx
dt
=
2
t
DE1
are the family of verticallyshifted parabolas shown to the right and given
explicitly below.
x
c
(
t
)
=
t
2
+
c
The critical curve is the vertical axis
t
=
0
. See the dotted black curve.
If we multiply
each of the above solutions by a multiplier
function
such as
m
(
t
)
=
e
−
t
we obtain the new family of functions
z
c
(
t
)
shown to the right and
defined below by:
z
c
(
t
)
≡
m
(
t
)
Multiplier
Function
⋅
x
c
(
t
)
=
e
−
t
⋅
t
2
+
c
(
)
The new critical curve is
z
=
e
−
t
⋅
2
t
. See the black curve.
Does this new family of functions also satisfy a differential equation? Yes. We
will verify that this new family of curves satisfies the new differential
equation:
dz
dt
+
z
=
e
−
t
⋅
2
t
DE2
Critical curve shown in black.
Using the product rule, the derivative of
z
is seen to be:
dz
dt
=
−
e
−
t
⋅
t
2
+
c
(
)
−
z
(
t
)
+
e
−
t
⋅
2
t
Recognizing the term
–
z
(
t
)
on the RHS, we bring it to the left and obtain:
dz
dt
+
z
=
e
−
t
⋅
2
t
In this case, the new differential equation DE2 appears harder
than the first one since DE1 has the form of a perfect
derivative:
dx
dt
=
f
(
t
)
. But we could easily run the above process in reverse. If we started with DE2, it can be
converted to the simpler DE1 by multiplying by the reciprocal of the multiplier. This is the idea behind Leibniz’s
multiplier method.
Another interesting question is how this process works in general. How can we find the new
differential equation that
results when the solutions to a given differential equation are multiplied by a multiplier function
m
(
t
)? The answer is
revealed in the following theorem.
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Multiplier Theorem
A.
Multiplier functions transform 1
st
order linear DEs to new 1
st
order linear DEs. Suppose the function
y
(
t
) is a
solution to the firstorder linear differential equation:
dy
dt
+
p
(
t
)
y
(
t
)
=
g
(
t
)
DE1
Next multiply each such solution by the multiplier
function
m
(
t
), obtaining the product function:
z
(
t
)
=
m
(
t
)
⋅
y
(
t
)
Then the product function
z
(
t
) is a solution to the new firstorder linear differential equation:
dz
dt
+
p
(
t
)
−
m
'(
t
)
m
(
t
)
⎛
⎝
⎜
⎞
⎠
⎟
P
(
t
)
⋅
z
(
t
)
=
m
(
t
)
⋅
g
(
t
)
G
(
t
)
DE2
B.
You can already see one potential application. The coefficient of
z
(
t
) can be eliminated
if we slyly choose the
multiplier function
m
(
t
) so that
p
(
t
)
−
m
'(
t
)
m
(
t
)
=
0
. This is exactly the idea behind Leibniz’s method. Rearranging the
above equation we find
dm
m
=
p
(
t
)
⋅
dt
for which one solution is Leibniz’s magic factor
m
(
t
)
=
e
p
(
t
)
⋅
dt
∫
.
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 Spring '08
 Hrebian
 Derivative, dt

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