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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 25m3
= 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 CauchyEuler Equation, so we expect the solution to have form y
=.
Computing derivatives, y' = m , y''
=m(m1) , y'''
= m(m1)(m3).
Plugging this in to the differential equation gives:
m(m1)(m3)2m(m1)

2x m +
8
= 0
m(m1)(m3)2m(m1)
2
m +8
= 0
m(m1)(m3)2m(m1)  2m + 8=0
 5
+ 2m + 8 = 0
(m+1)(m2)(m4) = 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
= (m1)(m1),
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=4x
For convergence, we need ρ = 4x < 1, so x
< 1/4, or 1/4 < x
< 1/4.
Check endpoint x = 1/4
==
(divergent pseries)
Check endpoint x = 1/4
==
(convergent alternating pseries)
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.