# solve differential equation HELP pls

• Mar 8th 2009, 11:25 AM
manalive04
solve 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 AM
HallsofIvy
Quote:

Originally Posted by manalive04
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

That sounds like a very good idea! What did you get?
• Mar 8th 2009, 11:59 AM
manalive04
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 PM
skeeter
Quote:

Originally Posted by manalive04
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?

$\displaystyle \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx$

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

can you integrate the LHS now?
• Mar 8th 2009, 12:12 PM
manalive04
Quote:

Originally Posted by skeeter
$\displaystyle \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx$

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

can you integrate the LHS now?

can you plese show me how im sorry im not confident here
• Mar 8th 2009, 12:19 PM
skeeter
hint: $\displaystyle \int \frac{u'}{u} \, dv$
• Mar 8th 2009, 12:22 PM
manalive04
Quote:

Originally Posted by skeeter
hint: $\displaystyle \int \frac{u'}{u} \, dv$

what is u'

im sorry i dont understand
• Mar 8th 2009, 12:37 PM
e^(i*pi)
Quote:

Originally Posted by manalive04
what is u'

im sorry i dont understand

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 PM
manalive04
Quote:

Originally Posted by e^(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

so what does this imply???

im so confused can u please make this clearer im sorry
• Mar 8th 2009, 01:51 PM
skeeter
$\displaystyle \frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx$

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

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