Well, the transfer function of the circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{sL}}{\text{sL}+\text{R}}\tag1$$

Where I used the well-known Laplace transform.

The impulse response of a system is given by the output response when a the input function is a dirac delta function, so:

$$\text{V}_\text{i}\left(\text{s}\right)=\mathscr{L}_t\left[\delta\left(t\right)\right]_{\left(\text{s}\right)}=1\tag2$$

The Laplace transform can be found using the definition of the Laplace transform or by examining a table of selected Laplace transforms.

Now, the output time response of this given by:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\underbrace{\text{V}_\text{i}\left(\text{s}\right)}_{=\space1}\cdot\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=-\frac{\text{R}}{\text{L}}\cdot\exp\left(-\frac{\text{R}t}{\text{L}}\right) \end{split}\tag3 \end{equation}

We can find the general output by using the convolution property of the Laplace transform:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\cdot\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int_0^t\text{v}_\text{i}\left(\tau\right)\cdot\left(-\frac{\text{R}}{\text{L}}\cdot\exp\left(-\frac{\text{R}\left(t-\tau\right)}{\text{L}}\right)\right)\space\text{d}\tau\\ \\ &=\frac{\text{R}}{\text{L}}\int_t^0\text{v}_\text{i}\left(\tau\right)\cdot\exp\left(\frac{\text{R}\left(\tau-t\right)}{\text{L}}\right)\space\text{d}\tau \end{split}\tag4 \end{equation}

So, for your case we get:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\frac{1\cdot1000}{1}\int_t^0\left(20\tau-5\right)\cdot\exp\left(\frac{1\cdot1000\left(\tau-t\right)}{1}\right)\space\text{d}\tau\\ \\ &=\frac{251\left(1-\exp\left(-1000t\right)\right)-1000t}{50} \end{split}\tag5 \end{equation}

Well, the transfer function of the circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{sL}}{\text{sL}+\text{R}}\tag1$$

Where I used the well-known Laplace transform.

The impulse response of a system is given by the output response when a the input function is a dirac delta function, so:

$$\text{V}_\text{i}\left(\text{s}\right)=\mathscr{L}_t\left[\delta\left(t\right)\right]_{\left(\text{s}\right)}=1\tag2$$

The Laplace transform can be found using the definition of the Laplace transform or by examining a table of selected Laplace transforms.

Now, the output time response of this given by:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\underbrace{\text{V}_\text{i}\left(\text{s}\right)}_{=\space1}\cdot\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=\frac{\partial}{\partial t}\left\{\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}+\frac{\text{R}}{\text{L}}}\right]_{\left(t\right)}\right\}\\ \\ &=\frac{\partial}{\partial t}\left\{\exp\left(-\frac{\text{R}t}{\text{L}}\right)\right\}\\ \\ &=-\frac{\text{R}}{\text{L}}\cdot\exp\left(-\frac{\text{R}t}{\text{L}}\right) \end{split}\tag3 \end{equation}

We can find the general output by using the convolution property of the Laplace transform:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\cdot\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t\right)}\\ \\ &=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{sL}}{\text{sL}+\text{R}}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int_0^t\text{v}_\text{i}\left(\tau\right)\cdot\left(-\frac{\text{R}}{\text{L}}\cdot\exp\left(-\frac{\text{R}\left(t-\tau\right)}{\text{L}}\right)\right)\space\text{d}\tau\\ \\ &=\frac{\text{R}}{\text{L}}\int_t^0\text{v}_\text{i}\left(\tau\right)\cdot\exp\left(\frac{\text{R}\left(\tau-t\right)}{\text{L}}\right)\space\text{d}\tau \end{split}\tag4 \end{equation}

So, for your case we get:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\frac{1\cdot1000}{1}\int_t^0\left(20\tau-5\right)\cdot\exp\left(\frac{1\cdot1000\left(\tau-t\right)}{1}\right)\space\text{d}\tau\\ \\ &=\frac{251\left(1-\exp\left(-1000t\right)\right)-1000t}{50} \end{split}\tag5 \end{equation}