Solve the following differential ecuation: $y'x^2 = 1+3y$ Solve the following differential equation: $y'x^2 = 1+3y$, with $y(3) = -\frac{1}3$
I tried the following but i'm a little bit lost at the end.

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


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


$\frac{1}{3}\ln(1+3y)=-\frac{1}{x}+c$


$\ln(1+3y)=-\frac{3}{x}+c$


$(1+3y)=e^{-\frac{3}{x}}e^c$


$y=\frac{(e^{-\frac{3}{x}}k)-1}{3}$

Well... here i dont know how to end the problem.
 A: You have done all the hard work.
Just put $x=3, y=-\frac13$
and solve for $k$.
A: This is the first thing you should check with separation, what to treat separately to  exclude division by zero. Here these are the values $x=0$ and $y=-\frac13$. Then establish that $y(x)=-\frac13$ is indeed a - constant - solution.
A: You correctly found that $$y = \frac{e^{-3/x}k - 1}{3} = k_2e^{-\frac{3}{x}} - \frac{1}{3}.$$
This is as good as it gets without an initial condition, i.e. $y(0) = \lambda$ for some $\lambda$, because when you solve a differential equation like this, you solve for a family of solutions. In this case, you are given that $y(3) = -1/3$, so $y(3) = k_2 e^{-3/3} - 1/3 = -1/3$ implies $k_2 = 0$.
A: You have one unknown variable ($k$) and one initial condition. Simply plug the IC into the general solution you've found.
$$y(3)=\frac{1}{3}(ke^{-3/3}-1)=-\frac{1}{3}$$
$$\Rightarrow ke^{-1}-1=-1$$
$$\Rightarrow ke^{-1}=0$$
$$\Rightarrow k=0$$
And thus you have the specific solution
$$y(x)=-\frac{1}{3}$$
A: 
$(1+3y)=e^{-\frac{3}{x}}e^c$

Start from here, let $~k=e^c~$ be some constant.
$$(1+3y)=k~e^{-\frac{3}{x}}$$
Plug in the initial condition $y(3)=-\frac{1}3$
$$\Rightarrow ~0=k~e^{-1}\Rightarrow k=0\Rightarrow 1+3y=0$$
Therefore, the final solution is:
$$y=-\frac{1}{3},~~~~x\in(0,\infty)$$
Note, since $x=0$ is a singularity of the original differential equation, the solution exists on $(-\infty, 0)$ or $(0, \infty)$. But your initial condition is at $x=3$, which means you choose the $(0, \infty)$ for the domain of $x$.
A: Alternatively, denote $u := 1 + 3 y$, so that the initial value problem becomes
$$\frac{u'}{x^2} = 3 u, \qquad u(3) = 0 .$$ Since the initial condition is $u(3) = 0$, we check and immediately find that $u(x) = 0$, $x > 0$, is a solution, so there's no need to integrate to find the general solution of the o.d.e. Solving for $y(x)$ gives $y(x) = -\frac{1}{3}$.
