Finding the solution to a differential equation

[jonnburton]

I've been working on this differential equation but haven't been able to find the correct solution and wondered whether anyone could point out where I'm going wrong.

$\displaystyle x\frac{dy}{dx} = y - y^2$

When y = 1/2, x =2

$\displaystyle \frac{1}{y -y^2}\frac{dy}{dx} = \frac{1}{x}$



$\displaystyle \frac{1}{y -y^2}$ needs to be split up into partial fractions before integration

$\displaystyle \frac{1}{y -y^2} = \frac{1}{y(1-y)}$

$\displaystyle \frac{A}{y} + \frac{B}{1-y}$ This comes out as $\displaystyle \frac{1}{y} + \frac{1}{1-y}$



Now it's possible to continue with the original question:

$\displaystyle \int \frac{1}{y}dy + \int \frac{1}{1-y}dy = \int \frac{1}{x}dx$

$\displaystyle ln|y| - ln|1-y| = ln|x| +A$


$\displaystyle e^{ln|y| - ln|1-y|} = e[{ln|x| +A}$

$\displaystyle \frac{y}{1-y} = Ae^{ln|x|} = Ax$

But the solution I came to when substituting in the values for x and y was $\displaystyle \frac{y}{1-y} = -\frac{1}{4}x$, whereas the correct solution is $\displaystyle y = \frac{x}{x+2}$

 

[Deleted member 4993]

I've been working on this differential equation but haven't been able to find the correct solution and wondered whether anyone could point out where I'm going wrong.

$\displaystyle x\frac{dy}{dx} = y - y^2$

When y = 1/2, x =2

$\displaystyle \frac{1}{y -y^2}\frac{dy}{dx} = \frac{1}{x}$



$\displaystyle \frac{1}{y -y^2}$ needs to be split up into partial fractions before integration

$\displaystyle \frac{1}{y -y^2} = \frac{1}{y(1-y)}$

$\displaystyle \frac{A}{y} + \frac{B}{1-y}$ This comes out as $\displaystyle \frac{1}{y} + \frac{1}{1-y}$



Now it's possible to continue with the original question:

$\displaystyle \int \frac{1}{y}dy + \int \frac{1}{1-y}dy = \int \frac{1}{x}dx$

$\displaystyle ln|y| - ln|1-y| = ln|x| +A$


$\displaystyle e^{ln|y| - ln|1-y|} = e[{ln|x| +A}$

$\displaystyle \frac{y}{1-y} = Ae^{ln|x|} = Ax$

But the solution I came to when substituting in the values for x and y was $\displaystyle \frac{y}{1-y} = -\frac{1}{4}x$, whereas the correct solution is $\displaystyle y = \frac{x}{x+2}$

You are messing up in algebra during substitution of initial values.

Ax = y/(1-y)

A*2 = 1

A = 1/2

y = Ax (1-y) = Ax - yAx

y (1+Ax) = Ax

y = Ax/(1+Ax)

y = 1/2 *x/(1+1/2*x)

y = (x/2)/[(2+x)/2] = x/(x + 2)

 

[jonnburton]

Thanks a lot for pointing that out, Subhotosh. I will go through it a few more times to make sure I get it.