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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)
=
(D-5)(D-3)
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 = 4-9
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(D-1) to both sides
D(D-1)(y''-2y'-3y) = D(D-1)(4-9)
D(D-1)(D-3)(D+1)y = 0
Auxiliary Equation:
m(m-1)(m-3)(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
= 4-9
B -
2(B
)
- 3(A + B
)
= 4
-9
-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[
|
|
] =![]() ![]() ![]() ![]() |
=∫
dx=-∫x
ln(x)dx=-
+∫
dx=-
+
=∫
dx=∫
ln(x)dx=x ln(x)-∫x
dx=x
ln(x)-x
Then: =(-
+
)
+(x
ln(x)-x)x
=-
+
+
ln(x)
-
=
-
SOLUTION
y=+
x
+
-