Question

Write y=ue^(ax) to find the general solution of y"-4y'+4y=e^(2x)/(1+x).

Answer 1

Just tell me the correct answer. Because I'm not getting the book's answer. I want to know the right answer.

Answer 2

@SithsAndGiggles

Answer 3

This one is very similar to the one you asked before.
You should get a=2 and u''=(1+x)^(-1)

Answer 4

http://www.wolframalpha.com/input/?i=y%27%27-4y%27%2B4y%3De%5E%282x%29%2F%281%2Bx%29
and I get the same thing as wolfram

Answer 5

you do know if you get \[\text{ blah blah }-x+C x \\ \text{ you can write this as } \\ \text{ blah blah }+(-1+C) x \\ \text{ which can be rewritten as } \\ \text{ blah blah } +c_1 x \\ \text{ similarly if you have } \\ \text{ blah blah } -1+D \\ \text{ you can write this as } \\ \text{ blah blah } +c_2\]

also maybe you problem is remembering how to integrate ln(1+x)?
you can do that by integration by parets.

Answer 6

let me know if you still any help on this problem
or like show me what you got have or whatever

Answer 7

Again, solve the complementary equation \(y''-4y'+4y=0\) first. Then make a guess on a particular solution. I would guess something like \(ce^{2x}\ln(1+x)\) since the first derivative of this is \(c\left(2e^{2x}\ln(1+x)+\dfrac{e^{2x}}{1+x}\right)\) and the second term is what we want.

Answer 8

Thank you!