If $\frac{dy}{dx} = -0.02y^2+0.2y$ and $y(0)>0$, what is $\lim_{x\to\infty}y(x)$? If $\frac{dy}{dx} = -0.02y^2+0.2y$ and $y(0)>0$, what is $\lim_{x\to\infty}y(x)$?
I thought that the answer was unsolvable since we don't know our value of y(0) but apparently the answer is 10 and I have no clue how to go about this
 A: The equation is
$$y'=-\frac 1 {50} y^2+\frac 15 y$$ Rewrite it as
$$\frac 1 {x'}=-\frac 1 {50} y^2+\frac 15 y\implies x'=\frac{50}{(10-y) y}=\frac{5}{y}+\frac{5}{10-y}$$ Integrate both sides
$$x+c=5\log\left(\frac y{10-y} \right)\implies c e^{\frac x 5}=\frac y{10-y}\implies y=\frac{10\, c\, e^{x/5}}{c\, e^{x/5}+1}$$ Using the condition $y(0)=a$ this gives
$$a=\frac{10\, c}{c\,+1}\implies c=\frac{a}{10-a}\implies y=\frac{10 a e^{x/5}}{a e^{x/5}+10-a}$$
A: This is called an "autonomous" equation, because $x$ doesn't appear.  You don't need to solve the equation to know it's long term behavior.  Just find the equilibrium points of the equation and determine their stability.  
Solve $-0.02y^2+0.2y =0$  to get $y=0$ and $y=10.$ These are the "equilibrium" points.   Since the derivative doesn't depend on $t$, once you know $y$ is $0$ (or $10$), then it's always $0$ (or $10$) because the derivative is zero there.  
Second, these points divide the $y$-axis into three regions:  $y<0$, $0 < y < 10$, and $y>10$.
In the first region,  $-0.02y^2+0.2y$ is negative, because $y$ is negative.  So the derivative is negative.  In the second region, the derivative is positive.  So that means that solutions near zero are pushed away from zero.  That is, negative solutions are decreasing and positive solutions are increasing.
Further, in the third region,  $-0.02y^2+0.2y$ is negative, so the solutions are decreasing.  That means solutions greater than $10$ decrease, while solutions less than $10$ increase.  So any solution that starts positive either decreases to $10$ or increases to $10$.  
$10$ is an "asymptotically stable" equilibrium point and $0$ is an "unstable" equilibrium point.
