If $\frac{dv}{dx} = y + 3 > 0$ and $y(0) = 2$, then $y(\ln 2)$ is equal to

There is likely a typo in the differential equation, and the intended equation is $\frac{dy}{dx}=y+3$, since the condition uses $y(0)$ and asks for $y(\ln 2)$. Then solve it as a first-order linear equation. $$\frac{dy}{dx}-y=3$$ A constant particular solution gives $y=-3$, and the complementary solution is $y=Ce^x$. So $$y=Ce^x-3$$ Using $y(0)=2$, $$2=C-3$$ $$C=5$$ Hence $$y=5e^x-3$$ Now evaluate at $x=\ln 2$: $$y(\ln 2)=5e^{\ln 2}-3$$ $$y(\ln 2)=5\cdot 2-3$$ $$y(\ln 2)=7$$ On checking the options, this matches option $\text{B}$.