Given differential equation

$\frac{dy}{dx}=y+3, \qquad y(0)=2$

We need to find:

$y(\ln 2)$

Rewrite in standard linear form

$\frac{dy}{dx}-y=3$

This is a first-order linear differential equation.

Find the integrating factor

For

$\frac{dy}{dx}+Py=Q,$

the integrating factor is

$IF=e^{\int P\,dx}$

Here, $P=-1$, so

$IF=e^{\int -1\,dx}=e^{-x}$

Multiply the equation by the integrating factor

$e^{-x}\frac{dy}{dx}-e^{-x}y=3e^{-x}$

The left side becomes:

$\frac{d}{dx}(ye^{-x})=3e^{-x}$

Integrate both sides

$ye^{-x}=\int 3e^{-x}\,dx=-3e^{-x}+C$

Multiply by $e^x$:

$y=-3+Ce^x$

Use the initial condition

Given $y(0)=2$:

$2=-3+Ce^0=-3+C$

$C=5$

So the solution is:

$y=5e^x-3$

Compute $y(\ln 2)$

$y(\ln 2)=5e^{\ln 2}-3=5\cdot 2-3=10-3=7$

Compare with options

The correct option is:

$\boxed{D: 7}$

Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.