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Lecture XVI
More Complicated Solutions to Differential Equations
I.
Nonlinear Differential Equations
A.
In fairly general terms, we can assert that a firstorder linear differential
equation has a unique solution.
This assertion follows our solution of the
&
()
xp
txg
t
+=
problem.
B.
However, solutions for nonlinear differential equations where
&
(, )
xf
t
x
=
are somewhat more difficult and lack a general solution framework.
For
example,
&
xx x
==
1
3
00
raises some additional difficulties.
C.
Two procedures for solving nonlinear differential equations:
1.
Separable Differential Equations
a.
Given the general expression of a differential equation:
dx
dt
ftx
=
an alternative way to express this relationship is
Mtx Ntx
0
.
One subgroup of this class of differential equations is referred
to as separable differential equations where
Mt Nx
( )
0
so that each term only involves t or x respectively.
Solving this
system then involves integrating each section separately:
Mt
Nx
Mtdt Nxd
x
M t dt
N x dx
=
=
=
∫∫
b.
Example 1:
tt
x
x
=
++

34
2
21
01
2
,()
Separating the function
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21
34
2
3
4
2
2
1
2
22
2
2
2
23
2
2
()(
)
()
(
)
xd
xt
td
t
zd
z
s
sd
s
c
xx
t
t
t
c
ttt
c
=+
+
=
+
++

=+
+
+
+
∫∫
Given that x(0) = 1,
121
3
=⇒=
cc
Thus, the implicit form of the solution is
2
2
3
+
+
+
c.
Example 2:
dx
dt
t
x
=
2
Separating the equation:
xdx t dt
z dz
s ds c
x
t
c
c
=
2
2
2
3
1
2
3
2.
Exact Differential Equations
a.
A second useful technique starts from a general equation
ψ
(, )
tx c
=
taking the total differential of this equation with dc = 0 we
obtain:
ψψ
tx
t x dt
t x dx
+=
0
0
or
Obviously, we see that since
ψ
s
t
z
x
t x dt
t x dx d
t x
sxd
s
tzd
z
=
=
∫
∫
or
Spotting this symmetry between the terms allows for the
solution of these problems.
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 Fall '08
 Moss

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