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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 First Order Linear Equations
Deﬁnitions. So far we have considered initial value problems in the form
dy
= −b + a y,
y (0) = y0
dt
where a and b are constants. We showed that the solution is
b
b
y (t) = + C eat
where
C = y0 − .
a
a
In general, an ordinary diﬀerential equation is said to be a ﬁrst order equation if it is a relation in the form
F (t, y, y ) = 0.
Upon expressing the derivative y in terms of the other variables, we ﬁnd an expression in the form
dy
= G(t, y )
dt
for some function G(t, y ). We say that a function y = y (t) is a solution if it satisﬁes this equation. If
y (t0 ) = y0 is an initial value, we think of the graph of y = y (t) as passing through the point (t0 , y0 ). We say
that this ﬁrst order equation is linear if G(t, y ) = g (t) − p(t) y for some functions g (t) and p(t). That is, if
the diﬀerential equation is in the form
dy
+ p(t) y = g (t).
dt
Integrating Factors. Consider the ﬁrst order linear diﬀerential equation
dy
+ p(t) y = g (t).
dt
Choose a function µ = µ(t). Multiply the ﬁrst order diﬀerential equation above by this function:
dy
+ µ(t) p(t) y = µ(t) g (t).
dt
Say that we can choose a function µ = µ(t) such that
i. µ(t) is not identically zero, and
ii. µ (t) = p(t) µ(t).
Then we would have
dy
dy dµ
d
µ(t)
+ µ(t) p(t) y = µ(t) ·
+
· y (t) =
[µ(t) y (t)]
dt
dt
dt
dt
This gives the diﬀerential equation
t
d
[µ(t) y (t)] = µ(t) g (t)
=⇒
µ(t) y (t) =
µ(τ ) g (τ ) dτ + C
dt
µ(t) for some constant C . (We use this notation to remind ourselves that the indeﬁnite integral of a function is
really a function of t, not the dummy variable of integration τ .)
Before we continue, we study conditions (i) and (ii) in more detail. If µ = µ(t) is not identically zero, we
can divide by it:
dµ
1 dµ
= p(t) µ
=⇒
= p(t).
dt
µ dt
The lefthand side is a derivative:
1 dµ
d
d
=
ln µ
=⇒
ln µ = p(t).
µ dt
dt
dt
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 Spring '09
 EdrayGoins
 Linear Equations, Equations

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