Given differential equation

$\frac{dp}{dt}=\frac12 p-200, \qquad p(0)=100$

Rewrite in standard linear form

$\frac{dp}{dt}-\frac12 p=-200$

This is a first-order linear differential equation.

Find the integrating factor

$\text{I.F.}=e^{\int -\frac12 dt}=e^{-t/2}$

Multiply the equation by the integrating factor

$e^{-t/2}\frac{dp}{dt}-\frac12 e^{-t/2}p=-200e^{-t/2}$

The left-hand side becomes:

$\frac{d}{dt}\left(pe^{-t/2}\right)=-200e^{-t/2}$

Integrate both sides

$pe^{-t/2}=\int -200e^{-t/2}\,dt + C$

Now,

$\int e^{-t/2}\,dt=-2e^{-t/2}$

so

$\int -200e^{-t/2}\,dt=400e^{-t/2}$

Hence,

$pe^{-t/2}=400e^{-t/2}+C$

Solve for $p(t)$

Multiply by $e^{t/2}$:

$p(t)=400+Ce^{t/2}$

Use the initial condition $p(0)=100$

$100=400+C$

$C=-300$

Therefore,

$p(t)=400-300e^{t/2}$

Match with the options

This is exactly Option C.

Verification

Differentiate:

$p'(t)=-150e^{t/2}$

Also,

$\frac12 p-200=\frac12(400-300e^{t/2})-200=200-150e^{t/2}-200=-150e^{t/2}$

So the differential equation is satisfied, and

$p(0)=400-300=100$

so the initial condition is also satisfied.