I went through this problem by adapting my work from a problem that had a constant volume, but my final answer is not realistic, so I know I messed up somewhere along the road. The question gives this information:
-Tank contains 200 liters of fluid in which 30 grams of salt is dissolved
-1 gram of salt per liter is pumped into the tank at a rate of 4L/min
-The well-mixed solution is then pumped out at a rate of 5L/min
From this, found the volume equation $V(t) = 200 - t$.
I let $x(t)$ be the amount of salt in grams dissolved in the tank.
I then set up the differential:
$$\frac{dx}{dt} = (1 g/L)(4L/min) - (5 L/min)\frac{x(t)}{200-t}.$$
I set up the equation in linear differential form:
$$\frac{dx}{dt} + \frac{5x(t)}{200-t} = 4.$$
I let $P(t) = \frac{5}{200-t}$.
So my integrating factor $\mu = e^{-5\ln(200-t)}$, which simplified was $\mu = (200-t)^{-5}$
My new equation is $x(t)(200-t)^{-5} = -(200-t)^{-4} + C$, which simplified is
$$x(t) = -(200-t)^1 + C(200-t)^5.$$
Using initial value $x(0) = 30$, I found $C = -170/200^5$
My final equation is $x(t) = -(200-t)-(170/200^5)(200-5)^5$.
Testing any number of $t$ values, I can tell this equation is nowhere near giving an accurate amount of salt in the tank, but I'm unsure which part of the problem I'm doing incorrectly.
You set the problem up and solved it perfectly except for ...
... a dropped factor of $-1$ on the RHS when you integrated.