# Finding A Second Solution For A DE Via Reduction Of Order

1. Oct 20, 2011

### Lancelot59

I'm given
$$y''-4y'+4y=0$$
and there is a solution $$y_{1}=e^{2x}$$
Using this I need to find a second solution.

Starting with the assumption:
$$y_{2}=u(x)*y_{1}$$
Then:
$$y=ue^{2x}$$
$$y'=2ue^{2x}+u'e^{2x}$$
$$y''=4ue^{2x}+2u'e^{2x}+u''e^{2x}+2u'e^{2x}$$

When I substitute back into the original equation, after doing the cancellation I wind up with:
$$u''e^{2x}=0$$
Then using the following order reduction:
$$w=u'$$
I get:
$$w'e^{2x}=0$$
I'm not sure what to do next here. The example in my textbook had two terms, and they used an integrating factor. How can I go about solving this problem? The answer given in the book is:
$$y_{2}=e^{2x}$$

2. Oct 20, 2011

### lurflurf

so
w' e2x=0
reduces to
w'=0
which is easy to solve
the answer given is wrong it should be
y2=(a x+b)e2x
where a and b are any numbers except a is not 0

3. Oct 20, 2011

### Lancelot59

I actually found a silly algebra error, so there should be two terms. However the answer is even stranger. I'll talk to my prof about it, thanks for looking!

4. Oct 21, 2011

### lurflurf

You were right the first time
w' e2x=0
reduces to
w'=0
notice when y is a solution so is y'-2y or conversely when
y'-2y=e2x
y is a solution of
y''-4y'+4=0

Last edited: Oct 21, 2011
5. Oct 21, 2011

### Lancelot59

I guess that does work. Thanks!

EDIT: I'm not sure how though. You have that w'=0, but how do you then solve for y being e^2x?

Last edited: Oct 21, 2011