Differential Equations, first order I am taking an Engineering Math class. When it comes of differential equations with $y^2$ included in the question I am lost on the procedure used to solve it. Bernoulli method does work because its not in form required and homogeneous equations doesn't work because of different degrees.

These are a few that had me stumped (Solving any one will be appreciated.)


$y' = (y^2 +9)/(xy+3y)$
$y \cdot \ln(x)y' = (y^2 +36) /x$
$y'=(x^2 y^4 - x^2) / y^3$


Thanks to anyone out there who can help.
For number one, here's a basic concept to help you get started:

$$y' = \frac{y^2 + 9}{xy + 3y} $$

$$y' = \frac{y^2 + 9}{y(x + 3)} $$

NOTE THE SEPARATION

$$\frac{y}{y^2+9}*\frac{dy}{dx} = \frac{y^2 + 9}{y(x + 3)}*\frac{y}{y^2+9}$$

$$\int \frac{y}{y^2+9}dy = \int \frac{1}{x+3} dx$$

$$\frac{\ln\left|{y^2+9}\right|}{2} = \ln\left|x+3\right| + C$$

$$\ln\left|{y^2+9}\right| = 2\ln\left|x+3\right| + C$$

$$y^2 + 9 =C(x+3)^2$$

Can you use this technique to solve the other problems? Comment if you have questions.
HINT 1

Start with the first one. Multiply across to get $(xy+3y)y'=y^2+9$. Now notice that the lhs has a factor $y$ and that $y y'$ is what you get if you differentiate $y^2$. What does that suggest?

Now look at the second one. Again we have a $y y'$. What happens if we try the same trick again?