Cells that are susceptible to HIV infection are called T (target) cells. Let $T(t)$ be the population of uninfected T-cells, $T*(t)$ that of the infected T-cells, and $V(t)$ the population of the HIV virus. A model for the rate of change of the infected T-cells is
$$\frac{dT^{ * }}{dt}=kVT-\delta T^{ * }, \quad (1)$$
where $\delta$ is the rate of clearance of infected cells by the body, and $k$ is the rate constant for the infection of the T-cells by the virus. The equation for the virus is the same as above:
$$\frac{dV}{dt}=P-cV, (2)$$
but now the production of the virus can be modeled by $P(t)=N\delta T^{ * }(t)$. Here $N$ is the total number of virions produced by an infected T-cell during its lifetime. Since $\frac{1}{\delta}$ is the length of its lifetime, $N\delta T^{ * }(t)$ is the total rate of production of $V(t)$.
At least during the initial stages of infection, $T$ can be treated as an approximate constant. Equations (1) and (2) are the two coupled equations for the two variables $T*(t)$ and $V(t)$.
A drug therapy using RT inhibitors blocks infection, leading to $k\cong 0$. Setting $k=0$ in (1), solve for $T*(t)$. Substitute it into (2) and solve for $V(t)$. Show the solution is $V(t)=\frac{V(0)}{c-\delta}[ce^{-\delta t}-\delta e^{-ct}]$.
I think the last paragraph of yours already gave you the direction. Set $k=0$ in (1), you will get
$$\frac{dT^{ * }}{dt}=-\delta T^{ * }$$
This can be solved by separation of variables. Plugging this solution into the second equation, you will get a linear differential equation. I'm sure you've learned how to solve these in differential equations.
Edit:
$$\frac{dV}{dt}=N\delta T^*(0)e^{-\delta t}-cV\\ \frac{dV}{dt}+cV=N\delta T^*(0)e^{-\delta t}$$
is a linear equation. You should multiply by $e^{ct}$ on both sides:
$$e^{ct}\frac{dV}{dt}+cVe^{ct}=N\delta T^*(0)e^{ct-\delta t}$$
which becomes
$$(e^{ct}V)'=N\delta T^*(0)e^{ct-\delta t}$$
Now can you integrate both sides and continue?