______________________________________________________________________
Differential Equations                              Test #2                                        April 18, 2005

Name____________________           R.  Hammack                                    Score ______
______________________________________________________________________

(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-

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[
] =-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 +-