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Differential Equations Test
#3 May
9, 2005
Name____________________ R. Hammack Score
______
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(1) Solve 2y''-3x
y'-3y=0 subject to the initial conditions y(1) = 1, y'(1)
= 0.
Auxiliary equation is 2-5m-3
= 0, or (2m + 1)(m - 3) = 0.
Hence, the general solution is y =+
.
y =+
y' =+3
From which we get:
1 =+
0 =+3
Or:
1 =+
0 = -+6
Adding,
1=7,
so
=1/7,
and
=6/7
SOLUTION: y =+
(2) Solve y'''-2
y''-2x y'+8y = 0.
This is a Cauchy-Euler Equation, so we expect the solution to have form y
=.
Computing derivatives, y' = m , y''
=m(m-1)
, y'''
= m(m-1)(m-3)
.
Plugging this in to the differential equation gives:
m(m-1)(m-3)
-2
m(m-1)
-
2x m
+
8
= 0
m(m-1)(m-3)-2m(m-1)
-2
m
+8
= 0
m(m-1)(m-3)-2m(m-1) - 2m + 8=0
- 5
+ 2m + 8 = 0
(m+1)(m-2)(m-4) = 0
Therefore the possible values for m are -1, 2, 4, and the solution to the differential
equation is:
y
=+
+
(3) Solve
y''-x y'+y = 2x.
First, let's find the complementary function.
y''- x y' + y = 0
Auxiliary equation is -2m+1
= (m-1)(m-1),
so =
x+
x
ln(x)
Now we continue, using variation of parameters.
Standard form is y'' -
y' +
y
=
Wronskian is Det(
|
|
) = x. |
Then =
∫
dx
= -2∫
dx
= -
Also =
∫
dx
=∫
dx
=2 ln(x)
so =
x
+
x
ln(x) = -
x
+2 ln(x)x ln(x) =
SOLUTION y =x
+
x
ln(x) +
(4) Find the
interval of convergence of the power series .
Using the ratio test for absolute convergence, we get
ρ==
=
=4|
x|
=4|x|
For convergence, we need ρ = 4|x| < 1, so |x|
< 1/4, or -1/4 < x
< 1/4.
Check endpoint x = 1/4
=
=
(divergent p-series)
Check endpoint x = -1/4
=
=
(convergent alternating p-series)
Conclusion: The interval of convergence is [-,
)
(5) Find find two linearly
independent power series solutions about the point x=0
of the differential equation y''-(1+x) y=0.
For solution, see Example 8 on page 245 of text.