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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)-2m(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 x ln(x) 1 ln(x)+1
) = 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.