I've got this RL circuit:
simulate this circuit – Schematic created using CircuitLab
And the Vin(t):
And I'd like to find Vout(t) using Laplace. I cannot figure out what the equation for this would look like so I'm here for help.
Thank you in advance :)
The KCL for the output node is:
$$\begin{align*} \frac{v_{_\text{OUT}}}{R_1}+\frac{1}{L_1}\int v_{_\text{OUT}}\,\text{d}t &= \frac{v_{_\text{IN}}}{R_1} \end{align*}$$
Taking the derivative to get rid of the integral,
$$\begin{align*} \frac{1}{R_1}\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+\frac{v_{_\text{OUT}}}{L_1} &= \frac{1}{R_1}\frac{\text{d}}{\text{d}t} v_{_\text{IN}} \\\\ \frac{L_1}{R_1}\frac{\text{d}}{\text{d}t}v_{_\text{OUT}}+v_{_\text{OUT}} &= \frac{L_1}{R_1}\frac{\text{d}}{\text{d}t} v_{_\text{IN}} \end{align*}$$
This is the same, in Laplace domain, as:
$$\begin{align*} \frac{L_1}{R_1}s\mathcal{L}\left\{v_{_\text{OUT}}\right\}+\mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \frac{L_1}{R_1}s\mathcal{L}\left\{v_{_\text{IN}}\right\} \\\\ \mathcal{L}\left\{v_{_\text{OUT}}\right\}\left(\frac{L_1}{R_1}s+1\right) &= \mathcal{L}\left\{v_{_\text{IN}}\right\}\left(\frac{L_1}{R_1}s\right) \\\\ \mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \mathcal{L}\left\{v_{_\text{IN}}\right\}\frac{\frac{L_1}{R_1}s}{\frac{L_1}{R_1}s+1} \\\\ \mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \mathcal{L}\left\{v_{_\text{IN}}\right\}\frac{s}{s+\frac{R_1}{L_1}} \end{align*}$$
Pause for a moment.
The first KCL equation above could have just been written straight out in Laplace form without the following steps (using \$V_{_\text{IN}}=\mathcal{L}\left\{v_{_\text{IN}}\right\}\$ and \$V_{_\text{OUT}}=\mathcal{L}\left\{v_{_\text{OUT}}\right\}\$) as \$\frac{V_{_\text{OUT}}}{R_1}+\frac{V_{_\text{OUT}}}{s L_1}=\frac{V_{_\text{IN}}}{R_1}\$ and then converted immediately to the latter equation above.
Moving on, you know that \$\mathcal{L}\left\{v_{_\text{IN}}\right\}=\mathcal{L}\left\{20\cdot t-5\right\}=\frac{20}{s^2}-\frac{5}{s}\$, right?
So,
$$\begin{align*} \mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \left(\frac{20}{s^2}-\frac{5}{s}\right)\cdot \frac{s}{s+\frac{R_1}{L_1}} \end{align*}$$
Surely, you can solve:
$$\begin{align*} \mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \left(\frac{20}{s^2}-\frac{5}{s}\right)\cdot \frac{s}{s+\frac{R_1}{L_1}} \\\\ \therefore \\\\ v_{_\text{OUT}}=\mathcal{L}^{-1}\mathcal{L}\left\{v_{_\text{OUT}}\right\} &= \mathcal{L}^{-1}\left\{\left(\frac{20}{s^2}-\frac{5}{s}\right)\cdot \left[\frac{s}{s+\frac{R_1}{L_1}}\right]\right\} \\\\ &=\mathcal{L}^{-1}\left\{\frac{20\frac{L_1}{R_1}}{s}-\frac{5\left(1+4\frac{L_1}{R_1}\right)}{s+\frac{R_1}{L_1}}\right\} \\\\ &=\mathcal{L}^{-1}\left\{\frac{20\frac{L_1}{R_1}}{s}\right\}-\mathcal{L}^{-1}\left\{\frac{5\left(1+4\frac{L_1}{R_1}\right)}{s+\frac{R_1}{L_1}}\right\} \\\\ &=20\frac{L_1}{R_1}\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}-5\left(1+4\frac{L_1}{R_1}\right)\mathcal{L}^{-1}\left\{\frac{1}{s+\frac{R_1}{L_1}}\right\} \end{align*}$$
You should be able to look these up in a table.
Jan and Jonk have already shown the way to solve this problem using Laplace transformation. However, when using Laplace a lot of (difficult) things are taken for granted. I will show a different approach to solving this problem, that doesn't involve Laplace which may peak the interest of OP and maybe some other on-lookers.
Writing up a node equation for \$V_\text{out} \$ gives us the differential equation: $$\frac{V_{\text{out}}}{R_1}-\frac{V_{\text{in}}}{R_1}+ \frac{1}{L_1}\int V_{\text{out}} \; \text{d}t=0 \Leftrightarrow$$
$$\frac{1}{R_1}\dot{V_\text{out}}+\frac{1}{L_1}V_\text{out}=\frac{\dot{V_\text{in}}}{R_1} \Leftrightarrow $$
$$\dot{V_\text{out}}+\frac{R_1}{L_1}V_\text{out} =\dot{V_\text{in}}$$
Observations:
Consequences
The decomposition principle holds for this LTIC system. This means, that we can find the total response of the system, as the sum of the system's zero-state response and zero-input response.
$$\text{Total response} = \text{zero-state response } + \text{ zero-input response} $$
The zero-state response is the output you measure when you set all initial conditions of the system to zero, and then apply your input signal (which in your case is a ramp). The output signal may be difficult to find, however, since the decomposition principle holds we can find the output as the convolution of the input signal with the system's impulse response.
The impulse response of a system is the output you measure when you apply an impulse \$\delta(t) \$ as input. In this case where \$V_\text{out}\$ and \$V_\text{in}\$ are differentiated of the same order, so the impulse response of the system can be found with this equation $$h(t) = b_1\delta(t) + \Big(P(D)y_n(t) \Big)u(t) $$ where \$b_1 \$ is the coefficient of \$\dot{V_\text{in}} \$, \$P(D) \$ is the differential operator for the right hand side of the diff. equation and \$u(t) \$ is a step function.
\$y_n(t) \$ is the natural response of your system, when the initial conditions of the system is set such that \$y_n^{(n-1)}=1 \$. In your case the natural response is found
$$ \dot{V_\text{out}}+1000V_\text{out} = 0, \: \: \: \: V_\text{out}(0^+) = 1 \Rightarrow $$
$$V_\text{out} = e^{-1000t} $$ So the impulse response becomes $$h(t) = \delta(t) -1000e^{-1000t}u(t) $$
As stated earlier, the zero-state response is found by convolving the input signal with the system's impulse response. $$y_\text{zs}(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau) d\tau $$ This can be done graphically and algebraically/computationally. In this case, it suitable to just use algebra and if you do that you get $$y_\text{zs}(t) = \Big(\frac{251}{50}(1-e^{-1000t})-5 \Big)u(t)$$
The zero-input response is the output you measure when there is no input signal, but the system is left to settle with its own initial conditions. In this problem, you haven't mentioned anything about initial conditions so one can assume that \$I_L(0^-) = 0 \$. And because the input voltage is likely a step function we can safely say that \$V_{out}(0^-) = 0 \$.
$$y_\text{zi}(t) = 0 $$
$$y_\text{total}(t) =y_\text{zs}(t) + y_\text{zi}(t) =\Big(\frac{251}{50}(1-e^{-1000t})-5 \Big)u(t) = \Big(0.02-5.02e^{-1000t}\Big)u(t) $$ And there is the total response of your system to the given input. Sorry for the long post, but I wanted to show how easily one can take things for granted. In one simple transformation, Laplace deals with all of the above concepts and computations in one elegant swoop. No differential equations, no integrals - just a variable "change" and some additional terms (if initial values are present).
To me, that is brilliance.
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}
