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Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 Upon integrating and exponentiating, we ﬁnd that
t
ln µ =
p(τ ) dτ + C1
=⇒ µ(t) = C2 exp t p(τ ) dτ where C2 = ±eC1 is some constant. This is the general solution to (ii). Since we want µ = µ(t) to satisfy
(i), we want C2 = 0. In fact, we will choose C2 = 1 – although this is not necessary. The function µ = µ(t)
is called an integrating factor.
We conclude the general solution of the original diﬀerential equation y = G(t, y ) in terms of G(t, y ) =
g (t) − p(t) y must be
t
t
1
y (t) =
µ(τ ) g (τ ) dτ + C
where
µ(t) = exp
p(τ ) dτ .
µ(t)
Example 1. We revisit the diﬀerential equation y = −b + a y for constants a and b. In this case, y + p(t) y =
g (t) where p(t) = −a and g (t) = −b are constant functions. The integrating factor would be
t
t
µ(t) = exp
p(τ ) dτ = exp
−a dτ = exp [−a t] = e−at .
Hence the general solution to the diﬀerential equation y = −b + a y must be
t
t
1
b
at
−at
at b −at
µ(τ ) g (τ ) dτ + C = e
−b e
dτ + C = e
e
+ C = + C eat .
y (t) =
µ(t)
a
a
Example 2. Say that we wish to solve the initial value problem
t y + 2 y = 4 t2 , y (1) = 2. First we ﬁnd the general solution to the diﬀerential equation, then we determine the constants of integration
from the initial condition.
Divide both sides of the equation by t so that the diﬀerential equation is in the form
dy
+ p(t) y = g (t)
dt
The integrating factor is
µ(t) = exp t where
p(τ ) dτ = exp p(t) =
t 2
,
t g (t) = 4 t.
2
dτ = exp [2 ln t] = t2 .
τ Multiply both sides of the diﬀerential equation y + p(t) y = g (t) by µ(t) = t2 : dy
+ 2 t y = 4 t3 .
dt
The lefthand side is the derivative of µ(t) y = t2 y , so we have
d2
C
t y (t) = 4 t3
=⇒
t2 y (t) = t4 + C
=⇒
y (t) = t2 + 2 .
dt
t
This is the general solution to the diﬀerential equation. For the initial condition, set t = 1:
1
1 + C = y (1) = 2
=⇒
C=1
=⇒
y (t) = t2 + 2 .
t
This is the solution to the initial value problem.
t2 First Order Nonlinear Equations
Deﬁnitions. Every ﬁrst order diﬀerential equation can be expressed in the form
dy
= G(t, y ).
dt
Recall that we say this equation is linear if G(t, y ) = g (t) − p(t) y for some functions p = p(t) and g = g (t).
We say that this equation is nonlinear otherwise. We say this ﬁrst order diﬀerential equation is separable if
we have the factorization G(t, y ) = Y (y ) T (t) in terms of functions Y = Y (y ) and T = T (t). ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue UniversityWest Lafayette.
 Spring '09
 EdrayGoins

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