Find a solution for IVP dy/dx=2x/(1-2y) , y(2)=0

Torinica

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Find the solution of the given initial value problem in explicit form"
dy/dx=2x/(1-2y) , y(2)=0

So I seperated to (1-2y)dy=2xdx, which I then integrated to get y-y2=x+c. I then found c to be -4.
So I have y-y2=x-4 but I have no idea how to get down to only one y.
Any help is appreciated!
 
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Find the solution of the given initial value problem in explicit form"
dy/dx=2x/(1-2y) , y(2)=0

So I seperated to (1-2y)dy=2xdx, which I then integrated to get y-y2=x+c. I then found c to be -4.
So I have y-y2=x-4 but I have no idea how to get down to only one y.
Any help is appreciated!

Why is \(\displaystyle \int\, 2y\, dy\, =\, y^2\) and \(\displaystyle \int\, 2x\, dx\, =\, x\)? [see red above]. Other than that, rewrite the equation as a quadratic in y and use the quadratic equation.
 
How do I rewrite it as a quadratic given two different variables?
How do you rewrite what as a quadratic? (And don't functions usually have [at least] two variables?)

When you reply, please show your work so far; in particular, please show the corrections you made as a result of the first reply. Then please clarify what it is that you're needing to do. For instance, does your book, when it says to "solve" a differential equation, expect an expression of y as a function of x, so y = f(x)? Thank you! ;)
 
How do I rewrite it as a quadratic given two different variables?
Since you used the small c as the constant of integration, I'll use capital letters for the quadratic equation:

A y2 + B y + C = 0

Now it doesn't make a bit of difference whether the A, B, and C are functions of y or of x or of both x and y, the solution to the equation is still given by the quadratic equation which is the solution to that problem. When A, B, and C are actual numbers, you can get a numeric solution but if A and/or B and/or C is a function of x, you would get a function y(x) as the solution.

So, as stapel suggested, please tell us your thoughts? What have you done so far? Show us your work even if you feel that it is wrong so we may try to help you. You might also read the thread you were asked to "Read Before Posting". Thank you!
 
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