Variation of Parameters/Wronskian

[smashyash]

So I'm doing some practice problems to prepare for a test on Friday and I'm just curious about this problem::

y'' + 3y' + 2y = 4e^(x)

in factoring using characteristics:

(r+2)(r+1) = 0
r = -2,-1

so Yc = C1*e^(-2x) + C2*e^(x)
y1= e^(-2x)
y2= e^(-x)

(skipping some algebra)..I used these to find u1' and u2':

u1' = -4e^(3x)
u2' = 4e^(2x)

Now, I thought that this was the part where you use y1, y1', y2, y2' do find the wronskian and divide u1' and u2' by w(y1,y2) but apparently no because the answer given in the back of the book tells me that the wronskian isn't even used... why is that? I thought the wronskian was included in the equation::

Yp = -y1(x) * INT( (y2(x)*f(x) ) / W(x) ) dx + y2(x) INT( (y1(x)*f(x) ) / W(x) ) dx

Maybe I'm not understanding this equation as well as I thought I did.. Any comments?? Thanks!

 

[timthereaper]

You use the Wronskian to find your u1' and u2':

u1' = - y2*f(x)/W
u2' = y1*f(x)/W

Then just integrate them to get the particular solutions.