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how would I integrate -y'/(y-3)y??
If you want to integrate that w.r.t. x(which I think is the case),I should say you can't do it. That should be a differential equation but if you don't specify what is it equal to,then its not even a differential equation and you can do nothing with it.For example you may say -y'/(y-3)y=y'' or any other thing and then you should go for finding a method for solving it.
Hi Mark!A different approach than the one Shyan gave is to use partial fraction decomposition to break up the left side.
You have
##\frac{dy}{y(3 - y)} = dx##
You can write the left side as [A/y + B/(3 - y)]dy and then integrate that with respect to y.
When I did the partial fractions business, I got this:
$$ \frac{1}{y(3 - y)} = \frac{1/3}{y} + \frac{1/3}{3 - y}$$
The last fraction is the same as -(1/3)/(y - 3). If you integrate, you get 1/3 [ ln|y| - ln|y - 3|] = 1/3 * ln[|y/(y - 3)|]. Notice that |3 - y| = |y - 3|.
The factor of 2 that you show in the original differential equation will change dx to 2dx, so the integral of the right side changes from x + C to 2x + C.
The initial DE is $$\frac{dy}{dx} = 2y(y-3)$$I think that you might have made a mistake when you exponentiated both sides (what you referred to as "taking the base e of both sides").
After I integrate, I get 1/3 * [ln|y| - ln|y - 3|] = 2x + C, or [ln|y| - ln|y - 3|] = 6x + C'. That's the same as ln[|y|/|y - 3|] = 6x + C'.
Without an initial condition, you can't get rid of the absolute values. Is there an initial condition that you haven't shown us?
Since
Two things:The initial DE is $$\frac{dy}{dx} = 2y(y-3)$$
After integrating I get
$$-ln|y-3|+ln|y| = 3(2x+C)$$
then after exponentiation I get..
$$\frac{y}{y-3} = e^(3(2x+C)) $$
|y - 3| and |3 - y| are identically equal, so you can change one for the other. The bigger question is what happened to the absolute values in the first place?$$\frac{y-3}{y} = \frac{1}{e^(3(2x+C))} $$
$$-\frac{3}{y} = \frac{1}{e^(3(2x+C))} -1$$
$$-\frac{3}{y} = \frac{1-e^(3(2x+C))}{e^(3(2x+C))} $$
$$y = -\frac{3e^(3(2x+C))}{1-e^(3(2x+C))} $$
But, via. the link provided, the solution is $$\frac{3e^(3(2x+C))}{e^(3(2x+C))+1}$$
since $$|y-3|$$ was changed to $$|3-y|$$
You can check to see. If they both satisfy the differential equation, they're both correct.Are both solutions correct?
*the 3(2x+C)'s are exponents
Two things:
1. Where did the absolute values go? As already mentioned, if there is an initial condition, you might be able to get rid of them, but you don't show an initial condition. Are you making an assumption that y(x) and y(x) - 3 are both positive?