This is a separable equation. We can rewrite the equation as it follows:
x\ y' -y = 3xy. Separating the x's and y's, we have that
xy' = y+3xy
xy' = (1+3x)y
{y'}/y = 1/x +3.
Now that variables are separated, we can integrate both sides with respect to x:
\int {y'}/y\ dx = \int ( 1/x +3)\ dx
Considering that y'\ dx = dy and that the integral of a sum is the sum of the integrals, we have
\int {dy}/y = \int 1/x\ dx + \int 3\ dx
Solving the ingrals, we have that
\log(y) = \log(x) + 3x+c
Solving for y:
y(x) = e^{\log(x) + 3x+c} = e^{\log(x)}\cdot e^{3x} \cdot e^c= x e^{3x} e^c.
To determine the value of c, we should use the condition y(1)=0 (if this is what you meant with y_1=0 in your question).
I'm a little puzzled with that one, because evaluating y for x=1 we have e^3 \cdot e^c, and this is never zero for c \in \mathbb{R}. I hope someone can correct me if I said something wrong, or that you can correct the question if the mistake was there.
