Help with a basic equation

**GIPC** · October 30th 2010, 08:38 AM

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.

**Jhevon** · October 30th 2010, 08:50 AM

Originally Posted by GIPC

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.

this is homogenious.

note that you have $\displaystyle \frac {dy}{dx} = \frac {2 \frac yx - 1}{2 - \frac yx}$

Let $\displaystyle v = \frac yx \implies y = vx \implies \frac {dy}{dx} = x \frac {dv}{dx} + v$

So your equation becomes:

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

Now that equation is separable. continue...

**GIPC** · October 30th 2010, 09:59 AM

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?

**Jhevon** · October 30th 2010, 10:07 AM

Originally Posted by GIPC

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?

you can't integrate like that!

It is a separable equation, meaning, you have to separate the variables. Only x's should be on one side, and only v's on the other. this includes the dx and dv

**GIPC** · October 30th 2010, 10:38 AM

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?

**Jhevon** · October 30th 2010, 11:36 AM

Originally Posted by GIPC

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?

Lets take it from $\displaystyle \ln \left| \frac {v - 1}{(v + 1)^3} \right| = 2 \ln |x| + C = 2 \ln |Ax| = \ln (Ax)^2$

Then we have

$\displaystyle \frac {v - 1}{(v + 1)^3} = (Ax)^2$

$\displaystyle \Rightarrow \frac {\frac yx - 1}{\left( \frac yx + 1 \right)^3} = (Ax)^2$

multiply the left side by $\displaystyle \frac {x^3}{x^3}$ to get

$\displaystyle \frac {x^2y - x^3}{(y + x)^3} = (Ax)^2$

divide both sides by $\displaystyle x^2$ to get

$\displaystyle \frac {y - x}{(y + x)^3} = A^2$

Plug in your initial data: $\displaystyle y(2) = 3$ and solve for $\displaystyle A$ . you end up with

$\displaystyle \frac {y - x}{(y + x)^3} = \frac 1{125}$

that's your solution, defined implicitly.

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