If $\dfrac{dy}{dx}=\dfrac{3x^2+5}{y}$ and $y=4$ when $x= 1$, then when $x=2$, $y=$

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$$\frac{dy}{dx}=\dfrac{3x^2+5}{y}$$ $$y   dy=3x^2+5   dx$$ $$\int y   dy= \int 3x^2+5   dx$$ $$\frac{1}{2}y^2=x^3+5x+C$$ $$y^2=2x^3+10x+C$$

Using our initial condition:

$$(4)^2=2(1)^3+10(1)+C$$ $$16=2+10+C$$ $$C=4$$

Rewriting and substituting $x=2$:

$$y^2=2(2)^3+10(2)+4$$ $$y^2 = 16+20 +4$$ $$y=\boxed{\pm\sqrt{40}}$$