Hello there,
L*di/dt is the voltage across the inductor:
v=L*di/dt
and that basically states that:
The voltage across the inductor is equal to the time rate of change of current (di/dt) through the inductor times the value of the inductance (L) of the inductor. That's not really enough information to analyze the circuit though because there is also a capacitor in the circuit and we have to know what that is doing too.
The best approach is probably to use a general circuit analysis procedure to solve for the peak voltage across the inductor. Using a general method means we would be able to solve this circuit as well as any other one.
It also helps to know a little about how the parts work in general too. For example, if the switch is closed for a very long period of time the inductor becomes a short circuit and so that absorbs all of the energy from the cap through the 10 ohm resistor (that got there when the switch was closed for the first time) and also all the energy from the battery. That means after the switch is closed and a very long time after that the capacitor voltage is zero and the inductor current is dependent only on the battery 5v and series resistor of 100 ohms. Since 5/100=0.050 that means for this circuit condition the inductor has 50ma flowing through it and it is a constant current so di/dt=0 so L*di/dt=0 so the voltage across the inductor is clearly zero volts. The circuit at this point basically looks like an inductor in series with a resistor in series with a battery.
Next, the switch is opened up. This means the next circuit looks like an inductor with initial current equal to 50ma in series with a resistor in series with a capacitor. Since the initial voltage across the capacitor is 0v and the initial current through the inductor is 0.050, that puts 50ma through the 10 ohm resistor. Now 50ma through 10 ohms is 0.050*10=0.5v, so that is the voltage across the inductor at the instant 0+ the switch is first opened. The circuit at that very instant 0+ is just a 10 ohm resistor in series with a current source of 50ma (the capacitor is a short circuit at 0+ too).
So you see we solved part of the problem already just knowing a little bit about how the inductor and capacitor behave during the first instant of operation and after very long time periods. To proceed however we would have to use a little general circuit analysis im not sure if you want to learn or not, but with two energy storage elements it's not super difficult.
We can get around using calculus and just use algebra by knowing some circuit analysis and using a table of Inverse Laplace Transforms and possibly some techniques. In this way we first do some algebra on the circuit in the frequency domain and then transform the algebra into the time domain using Inverse Laplace Transforms.
Here is a solution to your circuit after the switch is opened after it had been closed for a very long time and we ground the connection between the capacitor and inductor so the inductor voltage looks negative:
vL(t)=-e^(-a*t)*(A*sin(t*w)+B*cos(t*w))
where
a=R/(2*L)
w=sqrt(4*C*L-C^2*R^2)/(2*C*L)
A=(2*L-C*R^2)/(20*sqrt(4*C*L-C^2*R^2))
B=R/20
Also note that f0=w/(2*pi)=41.085923826
and also:
vL=-12.537 approximately at t=0.00586 seconds.
If we just want the voltage across the inductor non polarized then just make the above negative voltage positive.
I almost forgot, if you would like to try using numerical approximation techniques take a look here for a simple example of a capacitor being charged through a resistor with a voltage source:
https://www.electro-tech-online.com...ff-when-learning-capacitor.113051/#post925171