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 $\displaystyle \displaystyle \frac {dy}{dx} = \frac {2 \frac yx - 1}{2 - \frac yx}$
Let $\displaystyle \displaystyle v = \frac yx \implies y = vx \implies \frac {dy}{dx} = x \frac {dv}{dx} + v$
So your equation becomes:
$\displaystyle \displaystyle x \frac {dv}{dx} + v = \frac {2v - 1}{2 - v}$
Now that equation is separable. continue...
Lets take it from $\displaystyle \displaystyle \ln \left| \frac {v - 1}{(v + 1)^3} \right| = 2 \ln |x| + C = 2 \ln |Ax| = \ln (Ax)^2$
Then we have
$\displaystyle \displaystyle \frac {v - 1}{(v + 1)^3} = (Ax)^2$
$\displaystyle \displaystyle \Rightarrow \frac {\frac yx - 1}{\left( \frac yx + 1 \right)^3} = (Ax)^2$
multiply the left side by $\displaystyle \displaystyle \frac {x^3}{x^3}$ to get
$\displaystyle \displaystyle \frac {x^2y - x^3}{(y + x)^3} = (Ax)^2$
divide both sides by $\displaystyle \displaystyle x^2$ to get
$\displaystyle \displaystyle \frac {y - x}{(y + x)^3} = A^2$
Plug in your initial data: $\displaystyle \displaystyle y(2) = 3$ and solve for $\displaystyle \displaystyle A$. you end up with
$\displaystyle \displaystyle \frac {y - x}{(y + x)^3} = \frac 1{125}$
that's your solution, defined implicitly.