Question

Pls Help me find the general solution of (1+x) (dy/dx) - xy = x+ x^2

Answer 1

e^(x-ln|1+x|) y = integral of (e^x-ln|1+x|)) x

Answer 2

Then I'm stuck with the integral of something I don't quite see

Answer 3

your integrating factor is correct and the left hand side is also correct

Answer 4

\[\large \int e^{x - \ln |1+x|}x ~dx\]

Answer 5

are you stuck with that integral ?

Answer 6

Yes!

Answer 7

use exponent laws

Answer 8

\[\large a^{m-n} = \dfrac{a^m}{a^n}\]

etc

Answer 9

Then I just do integration by parts?

Answer 10

\[\large \int e^{x - \ln |1+x|}x ~dx = \int \dfrac{e^x}{e^{\ln |1+x|}}x ~dx = \int \dfrac{xe^x}{1+x} ~dx\]

Answer 11

yeah, maybe we can simplify a bit before IBP :
\[\large = \int \dfrac{(1+x-1)e^x}{1+x}~dx\]

Answer 12

\[\large = \int e^x - \dfrac{e^x}{1+x}~dx\]

Answer 13

not possible in elementary functions
http://www.wolframalpha.com/input/?i=%5Cint+e%5Ex%2F%281%2Bx%29+dx

so you're lucky, notthing to do here, just leave the integral as it is

Answer 14

Oh wow, thanks

Answer 15

wait a sec,
your integrating factor doesn't look correct :o

Answer 16

\[\large (1+x) \dfrac{dy}{dx} - xy = x+ x^2\]

Answer 17

\[\large \dfrac{dy}{dx} + \dfrac{ -x}{1+x}y = x\]

Answer 18

right ?

Answer 19

yes

Answer 20

thats what i have

Answer 21

IF = \[\large e^{\int \dfrac{-x}{1+x} ~dx} = e^{-\int \dfrac{1+x-1}{1+x} ~dx} \]

Answer 22

\[ \large e^{-\int 1 - \dfrac{1}{1+x} ~dx} = e^{-x + \ln |1+x|}\]

Answer 23

right ?

Answer 24

Notice that your signs were flipped in the exponent
thats the reason we arrived at a messedup integral

Answer 25

Oh I didnt carry the - sign

Answer 26

simplify the IF before multiplying both sides :

\[ \large e^{-\int 1 - \dfrac{1}{1+x} ~dx} = e^{-x + \ln |1+x|} = \dfrac{1+x}{e^x}\]

Answer 27

you may be able to integrate easily with this IF :) good luck !

Answer 28

Thanks so much!!

Answer 29

yeah wolfram says its integrable
http://www.wolframalpha.com/input/?i=%5Cint+x%281%2Bx%29%2Fe%5Ex+dx