dy/dx= (2x-y)/(x+y) all subject to y(1) = 1

I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help

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- Mar 8th 2009, 11:25 AMmanalive04solve differential equation HELP pls
dy/dx= (2x-y)/(x+y) all subject to y(1) = 1

I am not sure where to take this. I divided everything by x and made the subsequent y/x = V and then worked from there. please can you help - Mar 8th 2009, 11:49 AMHallsofIvy
- Mar 8th 2009, 11:59 AMmanalive04
i got an improper fraction which i am not sure what to do with

x(dv/dx) = (2 - 2v - v^2)/1+v

then took everything to the other side ==>

(1+v)/(2-2v-v^2)dv = dx/x

then to intergrate both sides

but how do you do the LHS? it is topheavy? - Mar 8th 2009, 12:06 PMskeeter
- Mar 8th 2009, 12:12 PMmanalive04
- Mar 8th 2009, 12:19 PMskeeter
hint: $\displaystyle \int \frac{u'}{u} \, dv$

- Mar 8th 2009, 12:22 PMmanalive04
- Mar 8th 2009, 12:37 PMe^(i*pi)
u is just a variable in this case.

$\displaystyle \int{\frac{u'}{u}}$ is a special case of integral and it equals $\displaystyle ln|x| + C$

it is saying that differentiating $\displaystyle 2-2v-v^2$ (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v - Mar 8th 2009, 12:44 PMmanalive04
- Mar 8th 2009, 01:51 PMskeeter
$\displaystyle \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx$

$\displaystyle \frac{-2 - 2v}{2-2v-v^2} \, dv = -\frac{2}{x} \, dx$

$\displaystyle \ln|2 - 2v - v^2| = -2\ln|x| + C$