Hello everyone,

I need some help with the following ordinary differential equation:

y'=(2y-x)/(2x-y), and y(2)=3

and I also have to solve it for y(0)=0 and explain what happens exactly.

I'll appreciate any help.

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- Oct 30th 2010, 09:38 AMGIPCHelp with a basic equation
Hello everyone,

I need some help with the following ordinary differential equation:

y'=(2y-x)/(2x-y), and y(2)=3

and I also have to solve it for y(0)=0 and explain what happens exactly.

I'll appreciate any help. - Oct 30th 2010, 09:50 AMJhevon
- Oct 30th 2010, 10:59 AMGIPC
let me see if I got it right.

I get

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

I integrate both sides and get

x*v=[(v^2-1)/(2-v)]*x +C

using v=y/x

y=...=[y^2-x^2]/(2x^2-xy) + C

now what do I do from here? - Oct 30th 2010, 11:07 AMJhevon
- Oct 30th 2010, 11:38 AMGIPC
how about this then

we reverse the sides and get

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

integrate and get

1/2[ln(1-v) - 3ln(v+1)]= lnx +C

so

ln[(1-y/x)/(y/x+1)^3]=2lnx+2c

(1-y/x)/[(y/x+1)^3]=x*e^2c

what shoud i do from here? - Oct 30th 2010, 12:36 PMJhevon