Solve the differential equation $\frac{dy}{dx} + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution Review 
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$; $x = 0$ when $y = 1$.

I solved this DE using the integration factor method. However, online calculators are giving me a different answer, where they instead used the separation of variables method.
Please review my solution and indicate if/where my reasoning is false, why it is false, how to fix it, and what the correct reasoning should be. Thank you.
My solution is as follows.
$\dfrac{dy}{dx} + 3xy = 0$
$e^{ \int 3x } dx = e^{ \frac{3x^2}{2} }$
$ \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3yxe^{ \frac{3x^2}{2} } = 0$
$ \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3xye^{ \frac{3x^2}{2}} $
$ \therefore \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = 0$
$ \displaystyle\int \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) dx + C = 0$
$ \Rightarrow y e^{ \frac{3x^2}{2}} + C = 0$
We want to solve for y. We know that $x = 0$.
$ \therefore ye^0 + C = 0$
$ \Rightarrow y = -C$

For the Initial Value Problem (IVP) $y′+3xy=0, ~y(0)=1$ (which you are most probably trying to solve) you are correct in all your steps above, except the last part: substitute $x=0, ~y=1$ and get $C=−1$, so the solution to the IVP is $y(x)=e^{−3x^2/2}.$
