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Differential Equations                                        Test #3                                                      May 9, 2005

Name____________________                   R.  Hammack                                                 Score ______
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(1)    Solve 2x^2y''-3x y'-3y=0 subject to the initial conditions y(1) = 1, y'(1) = 0.
Auxiliary equation is 2m^2-5m-3 = 0, or (2m + 1)(m - 3) = 0.
Hence, the general solution is y =c_1x^(-1/2)+c_2x^3.

y =c_1/x^(1/2)+c_2x^3
y' =-c_1/(2x^(1/2)^3)+3c_2x^2

From which we get:
1 =c_1+c_2
0 =-c_1/2+3c_2

Or:
1 =c_1+c_2
0 = -c_1+6c_2

Adding,
1=7c_2, so c_2=1/7, and c_1=6/7

SOLUTION:  y =6/(7x^(1/2))+x^3/7


(2)    Solve x^3y'''-2x^2 y''-2x y'+8y = 0.

This is a Cauchy-Euler Equation, so we expect the solution to have form y =x^m.
Computing derivatives,  y' = m x^(m - 1),   y'' =m(m-1) x^(m - 2),   y''' = m(m-1)(m-3)x^(m - 3).
Plugging this in to the differential equation gives:
x^3m(m-1)(m-3)x^(m - 3)-2x^2m(m-1) x^(m - 2)- 2x m x^(m - 1)+ 8x^m = 0
m(m-1)(m-3)x^m-2m(m-1) x^m-2 m x^m+8x^m = 0
m(m-1)(m-3)-2m(m-1) - 2m + 8=0
m^3 - 5m^2 + 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 =c_1x^(-1)+c_2x^2+c_3x^4

(3)    Solve x^2 y''-x y'+y = 2x.

First, let's find the complementary function.
x^2 y''- x y' + y = 0
Auxiliary equation is m^2-2m+1 = (m-1)(m-1),
so y_c=c_1x+c_2x ln(x)

Now we continue, using variation of parameters.
Standard form is y'' -1/x y' +1/x^2y =2/x

Wronskian is Det(
x x ln(x)
1 ln(x)+1
) = x.



Then u_1= ∫  (-x ln(x) 2/x)/xdx = -2∫  ln(x)/xdx = -(ln(x))^2
Also u_2= ∫  (x 2/x)/xdx =∫  2/xdx =2 ln(x)

so y_p=u_1x +u_2x ln(x) = -(ln(x))^2x +2 ln(x)x ln(x) = x(ln(x))^2

SOLUTION    y =c_1x + c_2x ln(x) + x(ln(x))^2

(4)    Find the interval of convergence of the power series Underoverscript[∑ , n = 1, arg3]4^n/n^(1/2)x^n.
Using the ratio test for absolute convergence, we get
ρ=Underscript[lim , n∞](| u_ (n + 1) |)/(| u_n |)=Underscript[lim , n∞](| 4^(n + 1)/(n + 1)^(1/2) x^(n + 1) |)/(| 4^n/n^(1/2) x^n |)=Underscript[lim , n∞](| x |^(n + 1))/(| x |^n)n^(1/2)/(n + 1)^(1/2)4^(n + 1)/4^n
=Underscript[lim , n∞]4| x|n/(n + 1)^(1/2)=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
Underoverscript[∑ , n = 1, arg3]4^n/n^(1/2)x^n=Underoverscript[∑ , n = 1, arg3]4^n/n^(1/2)(1/4)^n=Underoverscript[∑ , n = 1, arg3]1/n^(1/2) (divergent p-series)

Check endpoint x = -1/4
Underoverscript[∑ , n = 1, arg3]4^n/n^(1/2)x^n=Underoverscript[∑ , n = 1, arg3]4^n/n^(1/2)(-1/4)^n=Underoverscript[∑ , n = 1, arg3](-1)^n/n^(1/2) (convergent alternating p-series)

Conclusion: The interval of convergence is [-1/4, 1/4)


(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.