jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
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 \(\displaystyle y = \frac{x}{x+2}\)
\(\displaystyle x\frac{dy}{dx} = y - y^2\)
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 \(\displaystyle y = \frac{x}{x+2}\)