# Thread: solve differential equation HELP pls

1. ## 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

2. 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?

3. 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?

4. 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?
$\frac{1+v}{2-2v-v^2} \, dv = \frac{1}{x} \, dx$

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

can you integrate the LHS now?

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

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

can you integrate the LHS now?
can you plese show me how im sorry im not confident here

6. hint: $\int \frac{u'}{u} \, dv$

7. Originally Posted by skeeter
hint: $\int \frac{u'}{u} \, dv$
what is u'

im sorry i dont understand

8. Originally Posted by manalive04
what is u'

im sorry i dont understand
u is just a variable in this case.

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

it is saying that differentiating $2-2v-v^2$ (the denominator) gives -2-2v which is the numerator. The poster said that u = 2-2v-v^2 and thus u' = -2-2v

9. Originally Posted by e^(i*pi)
u is just a variable in this case.

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

it is saying that differentiating $2-2v-v^2$ (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

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

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

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