RL circuit unit impulse response
It's not complicated. You have two unknown voltage nodes:







simulate this circuit &ndash; Schematic created using CircuitLab

Note that I've made the two legs independent of each other, because they are. \$V_\text{IN}\cdot u_t\$ is just a fixed value, \$V_\text{IN}\$, multiplied by \$u_t\$ where \$\begin{align*}u_t&amp;\left\{\begin{array}{l}
t \lt0&amp; u_t=0\\t\ge0&amp;u_t=1\end{array}\right.\end{align*}\$.

So it should be pretty clear from the above that \$V_\text{X}\$ is independent of \$V_\text{Y}\$. So your questions can be treated, separately.

The initial conditions at \$t=0^-\$ will be that that current in \$L_1\$ is zero and that the voltage across \$C_1\$ will also be zero.

The nodal equations at \$t=0^+\$ are:

$$\begin{align*}
\frac{V_\text{X}}{R_1}+\frac{1}{L_1}\int V_\text{X}\:\text{d}t &amp;= \frac{V_\text{IN}}{R_1}&amp;\therefore\quad \frac{\text{d}V_\text{X}}{\text{d}t}+\frac{R_1}{L_1}V_\text{X}&amp;=0\:\text{A}\\\\
\frac{V_\text{Y}}{R_2}+C_1\frac{\text{d}V_\text{Y}}{\text{d}t}&amp;=\frac{V_\text{IN}}{R_2}&amp;\therefore\quad \frac{\text{d}V_\text{Y}}{\text{d}t}+\frac{1}{C_1\,R_2}V_\text{Y}&amp;=\frac{V_\text{IN}}{C_1\,R_2}
\end{align*}$$

The solutions, using the initial conditions of \$V_\text{X}=V_\text{IN}\$ at \$t=0^+\$ and \$V_\text{Y}=0\:\text{V}\$ at \$t=0^+\$, are:

$$\begin{align*}
V_\text{X} &amp;= V_\text{IN}\cdot e^{\frac{-t}{\left[\frac{L_1}{R_1}\right]}}\\\\
V_\text{Y} &amp;= V_\text{IN}\cdot \left(1-e^{\frac{-t}{R_2\,C_1}}\right)
\end{align*}$$

These are solved using the integrating factor for a 1st order ordinary differential equation.

To compute the values at \$t=1\:\text{ms}\$ you will need to have specific values for \$L_1\$, \$R_1\$, \$R_2\$, and \$C_1\$. You didn't provide those.

You may have questions about how to proceed with such general approaches. If so, feel free to ask.