______________________________________________________________________
Differential Equations                              Test #2                                        April 18, 2005

Name____________________           R.  Hammack                                    Score ______
______________________________________________________________________

(1) Given that y_1=x^3  is a solution to x^2y'' - 5x y' + 9y = 0, find a second, linearly independent solution.

First, let's put the equation in standard form: y''-5/xx y'+9/x^2y = 0

y_2=x^3∫e^(∫ -5/x  dx)/(x^3)^2dx =x^3∫e^(5 ln x  dx)/x^6dx =
x^3∫x^5/x^6dx =x^3∫1/xdx = x^3ln(x)



(2) Find a differential operator (of lowest possible degree) that annihilates the function y=3+x^2+x e^x.

D^3knocks out 3+x^2
(D - 1)^2knocks out x e^x

Thus the operator we seek is D^3(D - 1)^2 = D^3(D^2- 2D + 1) = D^5-2D^4+D^3




(3)  Find four linearly independent functions that are annihilated by the differential operator  D^4 - 8D^3 + 15D^2

  D^4-8D^3+15D^2 = D^2(D^2-8D+15) = D^2(D-5)(D-3)

The functions are thus:
y = 1
y = x
y = e^x
y = x e^x


(4)
(a)  Solve y''-4y'-5y = 0

Auxiliary Equation:
m^2- 4m - 5 = 0
(m - 5)(m + 1) = 0

Solution: y=c_1e^(5x)+c_2e^(-x)



(b) Find the solution to  y''-4y'-5y = 0 that satisfies y(0) = 3  and y '(0) = 5

y = c_1e^(5x)+c_2e^(-x)
y' = 5c_1e^(5x)-c_2e^(-x)

From this, we get

3 = c_1e^0+c_2e^0
5 = 5c_1e^0-c_2e^0

Or
3 = c_1+c_2
5 = 5c_1-c_2

Adding:
8 = 6c_1
c_1 = 4/3

Also,
3 = 4/3 +c_2
c_2 = 5/3

Thus, solution is y = 4/3e^(5x)+5/3e^(-x)



(5)   Solve  y''-2y'-3y = 4e^x-9

First let's find the complementary function:
y''-2y'-3y = 0
Auxiliary Equation:
m^2- 2m - 3 = 0
(m - 3)(m + 1) = 0
Thus: y_c=c_1e^(3x)+c_2e^(-x)

Now, back to the original equation.
y'' - 2y' - 3y = 4e^x - 9
Apply D(D-1) to both sides
D(D-1)(y''-2y'-3y) = D(D-1)(4e^x-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=c_1e^(3x)+c_2e^(-x)+ A + B e^x

y_p = A+B e^x
y_p' = B e^x
y_p'' = B e^x

Plugging this back into y''-2y'-3y = 4e^x-9
B e^x- 2(B e^x) - 3(A + B e^x) = 4e^x-9
-4B e^x- 3A = 4e^x- 9
B = -1
A = 3

SOLUTION
y=c_1e^(3x)+c_2e^(-x)+3-e^x


(6)  Solve y''+2y'+y = e^(-x)ln(x)

First let's find the complementary function:
y''+2y'+y = 0
Auxiliary Equation:
m^2+ 2m +1 = 0
0=(m+1)(m+1)
Thus: y_c = c_1e^(-x)+c_2x e^(-x)

Wronskian = Det[
e^(-x) x e^(-x)
-e^(-x) e^(-x) - x e^(-x)
] =e^(-2x)-x e^(-2x)+x e^(-2x)=e^(-2x)



u_1=∫(-x e^(-x) e^(-x) ln(x))/e^(-2x)dx=-∫x ln(x)dx=-(x^2ln(x))/2+∫x^2/21/xdx=-(x^2ln(x))/2+x^2/4

u_2=∫(e^(-x) e^(-x) ln(x))/e^(-2x)dx=∫ ln(x)dx=x ln(x)-∫x1/xdx=x ln(x)-x

Then: y_p=(-(x^2ln(x))/2+x^2/4)e^(-x)+(x ln(x)-x)x e^(-x)
y_p=-(x^2ln(x) e^(-x))/2+x^2/4e^(-x)+x^2 ln(x)e^(-x)-x^2e^(-x)
y_p=(x^2ln(x) e^(-x))/2-(3x^2)/4e^(-x)


SOLUTION
y=c_1e^(-x)+c_2x e^(-x)+(x^2ln(x) e^(-x))/2-(3x^2)/4e^(-x)