IVP: $y'=\frac{y}{3x-y^2}$, $y(1)=1$ Solve implicitly the initial value problem:
$
\left\{
  \begin{array}{l l}
    y'=\frac{y}{3x-y^2} & \quad x\geq 1\\
    y(1)=1
  \end{array} \right.
$
The equation is not exact and trying to transform it to one by integrating factor did not work. What else can I do?
 A: Remember that a integrating factor is a function $\mu$ such that $\mu P \ \mathrm{d}x + \mu Q \ \mathrm{d}y = 0$ is exact, that is: $$\mu \frac{\partial P}{\partial y} + P \frac{\partial \mu}{\partial y} = \mu \frac{\partial Q}{\partial x} + Q \frac{\partial \mu}{\partial x}$$
If $\mu$ happens to be only a function of $y$, we can write $\frac{\partial \mu}{\partial x} \equiv 0$ and $\frac{\partial \mu}{\partial y} = \mu'$. Rearranging, we get: $$\frac{\mu'}{\mu} = \frac{\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}}{P}$$
Checking the right-hand side of the above is the test to see if $\mu$ is really a function of $y$, only.
Write your equation as: $$ - y  \ \mathrm{d}x + (3x - y^2)\  \mathrm{d}y = 0$$
Calling $P = -y$ and $Q = 3x - y^2$. We have that: $$\frac{\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}}{P} = \frac{3 - (-1)}{-y} = -\frac{4}{y}$$
So your integrating factor is a function of $y$, and we have: $$\frac{\mu'(y)}{\mu(y)} = -\frac{4}{y}$$
Off you go (:
