- #1
Lancelot59
- 646
- 1
I'm given
[tex]y''-4y'+4y=0[/tex]
and there is a solution [tex]y_{1}=e^{2x}[/tex]
Using this I need to find a second solution.
Starting with the assumption:
[tex]y_{2}=u(x)*y_{1}[/tex]
Then:
[tex]y=ue^{2x}[/tex]
[tex]y'=2ue^{2x}+u'e^{2x}[/tex]
[tex]y''=4ue^{2x}+2u'e^{2x}+u''e^{2x}+2u'e^{2x}[/tex]
When I substitute back into the original equation, after doing the cancellation I wind up with:
[tex]u''e^{2x}=0[/tex]
Then using the following order reduction:
[tex]w=u'[/tex]
I get:
[tex]w'e^{2x}=0[/tex]
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:
[tex]y_{2}=e^{2x}[/tex]
[tex]y''-4y'+4y=0[/tex]
and there is a solution [tex]y_{1}=e^{2x}[/tex]
Using this I need to find a second solution.
Starting with the assumption:
[tex]y_{2}=u(x)*y_{1}[/tex]
Then:
[tex]y=ue^{2x}[/tex]
[tex]y'=2ue^{2x}+u'e^{2x}[/tex]
[tex]y''=4ue^{2x}+2u'e^{2x}+u''e^{2x}+2u'e^{2x}[/tex]
When I substitute back into the original equation, after doing the cancellation I wind up with:
[tex]u''e^{2x}=0[/tex]
Then using the following order reduction:
[tex]w=u'[/tex]
I get:
[tex]w'e^{2x}=0[/tex]
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:
[tex]y_{2}=e^{2x}[/tex]