If you do it directly, yes, the Wronskian is a big mess, especially because you are taking second derivatives of products. But if you use complex numbers the whole thing becomes more managable:
First, write e-2xcos(x sqrt(5)) as
Re(e-2x+i sqrt(5)x)
where Re is the real part. The imaginary part is the other function.
Next, to feel like the notation is not getting the best of us, we write a=2, b=-2+i sqrt(5), so that you're getting the Wronskian of
(eax, Re ebx, Im ebx).
The derivative is
(a eax, Re b ebx, Im b ebx).
and 2nd deriv. is
(a2 eax, Re b2 ebx, Im b2 ebx).
The resulting determinant, if we expand along the first column, involves
determinants of the form
Re bkebx Im bkebx
Re bnebx Im bnebx
which is of the form
Re c Im c
Re d Im d
which is Re c Im d + Re d Im c.
This is the imaginary part of cd.
So these determinants are of the form
Im (bk+n e2bx)
The overall Wronskian, then, is
eax Im(b3 e2bx)-aeax Im(b2 e2bx)+a2eax Im(b e2bx)
In this case, a = 2, so we have
e2x Im ((b3+2b2+4b)e2bx)
Now since b=-2+i sqrt(5), we compute
b3-2b2+4b
= (22+7 sqrt(5)i)-2(-1-4 sqrt(5)i)+4(-2 + sqrt(5)i)
=16+19 sqrt(5)i
and we get that the Wronskian is
e2x Im (16+19 sqrt(5)i)e-4x+2 sqrt(5)ix
= e2x (16e-4xsin(2 sqrt(5)x)+19 sqrt(5)e-4xcos(2 sqrt(5)x))
= 16e-2xsin(2 sqrt(5)x)+19 sqrt(5)e-2xcos(2 sqrt(5)x)
Kevin |
