LBS 119  EXAM #3 – Review Topics
1.
SEPARABLE EQUATIONS (1
st
order):
•
Arrange the equation so that
dy
dx
is on the left side of the equation by itself.
•
Separate the x’s and y’s on the other side by factoring out to get
( ) ( )
dy
f x g y
dx
=
.
•
Bring the y’s to the left side and x’s to the right side to get
1
( )
( )
dy
f x dx
g y
=
.
•
Integrate both sides of the equation.
•
If possible, solve for y.
2.
LINEAR EQUATIONS (1
st
order):
•
Put in standard form:
( )
( )
dy
P x y
Q x
dx
+
=
.
•
Find the integrating factor:
( )
P x dx
e
μ
=
•
Multiply the differential equation by the integrating factor:
( )
( )
dy
P x y
Q x
dx
+
=
•
The left side becomes
[
]
y
so we get
[
]
( )
y
Q x
=
•
Integrate both sides to get
( )
y
Q x dx C
=
+
•
Divide through by
to finish solving for y.
3.
BERNOULLI EQUATIONS (1
st
order):
•
Put in standard form:
( )
( )
n
dy
P x y
Q x y
dx
+
=
•
Let
1
(1
)
n
n
dv
dy
v
y
and
n y
dx
dx


=
=

.
•
Multiply the standard form equation by
(1
)
n
n y


and substitute to get a standard
form LINEAR equation with v’s instead of y’s.
•
Finish solving for v with the LINEAR method.
•
Substitute y back into the equation and solve for y if possible.
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View Full Document4. EXACT EQUATIONS
(Remember, these involve PARTIAL derivatives):
•
Put into form
( , )
( , )
0
M x y dx
N x y dy
+
=
and CHECK for EXACTNESS:
M
N
y
x
x
=
•
If exact, then there exists a function
( , )
f x y
such that
f
f
M and
N
x
y
x
=
=
(this is by definition).
Start with either one of these facts and work through
to find
( , )
f x y
using partial
integrals and partial
derivatives.
Remember to be
extra careful when keeping track of which variable is considered a variable and
which variable is considered a constant in each step.
•
The general solution is
( , )
f x y
C
=
5.
HOMOGENEOUS LINEAR EQUATIONS (2
nd
order or higher):
Example:
ay” + by’ + cy =0.
Use a characteristic equation to solve these.
An n
th
order equation needs n linearly
independent parts to its general solution.
Remember, to have linear independence, no
two parts of the solution can be constant multiples of each other.
If the characteristic equation has
•
DISTINCT REAL ROOTS
r
1
and r
2
, the general solution is
1
2
1
2
=
+
r x
r x
c
y
C e
C e
•
REPEATED REAL ROOTS
r
1
= r
2
, the general solution is
1
1
1
2
=
+
r x
r x
c
y
C e
C xe
•
COMPLEX or IMAGINARY ROOTS
1
2
α
β
=
+
=

r
i
and
r
i
,
the general solution is
1
2
cos
sin
=
+
x
x
c
y
C e
x C e
x
6.
NONHOMOGENEOUS LINEAR EQUATIONS (2
nd
order or higher):
Example:
y” + by’ + cy = g(x).
(make sure there is no leading coefficient!)
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 Spring '08
 HanniNichols
 Derivative, Elementary algebra, dy

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