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Differential Equations Test
#2 April
18, 2005
Name____________________ R. Hammack Score
______
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(1) Given that = is
a solution to y''
 5x y' + 9y = 0, find a second, linearly independent solution.
First, let's put the equation in standard form: y''x
y'+y
= 0
=∫dx
=∫dx
=
∫dx
=∫dx
= ln(x)
(2) Find a differential operator (of
lowest possible degree) that annihilates the function y=3++x
.
knocks
out 3+
knocks
out x
Thus the operator we seek is
= (
2D + 1) = 2+
(3) Find four linearly independent
functions that are annihilated by the differential operator
 8
+ 15
8+15
= (8D+15)
= (D5)(D3)
The functions are thus:
y = 1
y = x
y =
y = x
(4)
(a) Solve y''4y'5y = 0
Auxiliary Equation:

4m  5 = 0
(m  5)(m + 1) = 0
Solution: y=+
(b) Find the solution to y''4y'5y
= 0 that satisfies y(0) = 3 and y '(0)
= 5
y = +
y' = 5
From this, we get
3 = +
5 = 5
Or
3 = +
5 = 5
Adding:
8 = 6
= 4/3
Also,
3 = 4/3 +
= 5/3
Thus, solution is y = +
(5)
Solve y''2y'3y = 49
First let's find the complementary function:
y''2y'3y = 0
Auxiliary Equation:

2m  3 = 0
(m  3)(m + 1) = 0
Thus: =+
Now, back to the original equation.
y''  2y'  3y = 4
 9
Apply D(D1) to both sides
D(D1)(y''2y'3y) = D(D1)(49)
D(D1)(D3)(D+1)y = 0
Auxiliary Equation:
m(m1)(m3)(m+1) = 0
m = 3, m = 1, m =1, m = 0
y=++
A + B
= A+B
'
= B
''
= B
Plugging this back into y''2y'3y
= 49
B 
2(B )
 3(A + B )
= 49
4B 
3A = 4
9
B = 1
A = 3
SOLUTION
y=++3
(6) Solve
y''+2y'+y = ln(x)
First let's find the complementary function:
y''+2y'+y = 0
Auxiliary Equation:
+
2m +1 = 0
0=(m+1)(m+1)
Thus:
= +x
Wronskian = Det[


] =x +x = 
=∫dx=∫x
ln(x)dx=+∫dx=+
=∫dx=∫
ln(x)dx=x ln(x)∫xdx=x
ln(x)x
Then: =(+)+(x
ln(x)x)x
=++
ln(x)
=
SOLUTION
y=+x
+