Hi, can someone please check my work. I only have one chance to submit my answer online and I wanna be 100% sure.

$\displaystyle \frac{dx}{dy} + 3y = sin(x) , y(0) = 4$

$\displaystyle e^3 dx = e^{3x}$

$\displaystyle e^{3x} \frac{dx}{dy} + e^{3x}y = e^{3x}sin(x) dx$

$\displaystyle \int\frac{d}{dx} e^{3x}y = \int e^{3x}sin(x)$

$\displaystyle e^{3x}y = \frac{e^{3x}3sin(x)-cos(x)}{10} + c $

$\displaystyle y = \frac{3 sin(x) - cos(x)}{10} $