find the solution of differential equation that passes through the indicating points Find the solution of differential equation that passes through the indicating points 
 $dy/dx-y^2 = -9$  (0,3)
I have tried to solve it
$dy/dx = -9 + y^2$
$dy/dx = (y)^2 - (3)^2$
$dy/dx= (y-3).(y+3)$
$dy= (y-3).(y+3) dx$
What next ?
 A: There is a simple way of solving any equation of the type $y'= f(x)g(y)$. There is some cheating in the notation in the derivation that follows, however it is the quickest way to memorize it, so as long as you know where you are going, it is acceptable to do the steps outlined.
$$\frac{dy}{dx} = f(x)\cdot g(y)\text{ ("multiply" by }dx, \text{ divide with }g(y))\\
\frac{dy}{g(y)}= \frac{dx}{f(x)} \text{ (integrate)}\\
\int\frac{dy}{g(y)} = \int\frac{dx}{f(x)} + C
$$
This way, you get two functions, $F$ and $G$, and an equation $G(y) = F(x) + C$. Solving it for $y$ gives you the general form for the solution.
A: I would use this method for solving it
1) take everything apart from dy/dx on to the right hand side 
2) multiply both sides by dx and divide both sides by y^2-9 (should leave one dx on the right and one dy on the left hand side of the equality)
3) integrate both sides (don't forget to include constant C on the right hand side), you need to use the integral that tackles 1/(y^2 - a^2), it will give you a 1/2a*ln{y-a/y+a} (look it up) - in this case a=3 
4) set the left hand side of what you have (so the integral of 1/(y^2 - a^2))to u and find the inverse
5) insert the right hand side (should have integrated 1 to get "x + C") into to this new function of u (it should be y(u)) and that is your solution
Then just check to make sure that the points lie on the solution 
