# Help with a basic equation

• Oct 30th 2010, 09:38 AM
GIPC
Help 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 AM
Jhevon
Quote:

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

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

Now that equation is separable. continue...
• Oct 30th 2010, 10:59 AM
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?
• Oct 30th 2010, 11:07 AM
Jhevon
Quote:

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
• Oct 30th 2010, 11:38 AM
GIPC

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 PM
Jhevon
Quote:

Originally Posted by GIPC

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

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