My book doesn't explain this thoroughly, but apparently I'm supposed to start by solving
$$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x+3y}{x}$$
But is the rest correct?
$$\mathrm{d}y = \dfrac{2x+3y}{x}\mathrm{d}x \implies y = 2x+3y\ln x+C \implies y = \dfrac{2x+C}{1-3\ln x}$$
at $(4,5)$:
$$5 = \dfrac{2(4)+C'}{1+3\ln (4)} \implies C' = \dfrac{5+30\ln 2}{8} \implies y = \dfrac{2x+C'}{1-3\ln x}$$
Thanks in advance.
Hint
This does not seem to be correct.
$$\frac{dy}{dx} = \dfrac{2x+3y}{x}\implies \frac{dy}{dx}-3\frac{y}x =2$$ Solve the homegeneous equation first (it is quite simple). Then, use the variation of parameters to take into account the rhs.
I am sure that you can take it from here.