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Differential Equations                                        Quiz #8                                                      April 29, 2005

Name____________________                   R.  Hammack                                                 Score ______
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(1)    Find the general solution of the differential equation  x y''+y'=x

First, let's find y_c
x y''+y'=0
x ^2y''+x y'=0
A.E.:  m^2=0
y_c=c_1+c_2ln(x)

Next, apply variation of parameters to find y_p.
Standard Form:   y''+1/xy' = 1

W=Det(
1 ln(x)
0 1/x
) =1/x


u_1=  (-ln(x) 1)/(1/x)  dx=-∫x ln(x)dx = -(x^2ln(x))/2+x^2/4 (by parts)
u_2=  1/(1/x)  dx=∫x dx =x^2/2
Thus y_p=-(x^2ln(x))/2+x^2/4+x^2/4ln(x) = x^2/4

SOLUTION:
y=c_1+c_2ln(x) + x^2/4


(2)  Find the interval of convergence of the power series Underoverscript[∑ , n = 2, arg3](-2)^n/nx^n

Ratio Test for Absolute Convergence:
Underscript[lim , n∞](| (-2)^(n + 1)/(n + 1) x^(n + 1) |)/(| (-2)^n/nx^n |) = Underscript[lim , n∞](2^(n + 1) n | x |^(n + 1))/(2^n (n + 1) | x |^n) = Underscript[lim , n∞]2n/(n + 1)|x| = 2|x|
Thus, we get convergence if 2|x| < 1, or rather if -1/2 < x < 1/2.

What about the endpoint x = 1/2 ?  Then the series becomes  Underoverscript[∑ , n = 2, arg3](-2)^n/n(1/2)^n=Underoverscript[∑ , n = 2, arg3](-1)^n/nwhich is the (convergent) alternating harmonic series.

What about the endpoint x = -1/2 ?  Then the series becomes  Underoverscript[∑ , n = 2, arg3](-2)^n/n(-1/2)^n=Underoverscript[∑ , n = 2, arg3]1/nwhich is the (divergent)  harmonic series.

CONCLUSION: Interval  of convergence is (-1/2,1/2]


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