And then we can write that isc,

now this is KCL at this node, is equal to "iR" plus "i".

Now "iR" is the voltage, v,

over Rth and just keep that i.

The next thing we do is substitute v = L di/dt in for the v here.

So we'll have

iSC = I/Rth (L

di/dt) + i.

So at this point, we have a differential equation in terms of i, but

it remains to put it into standard form.

So to do that we isolate the di/dt term,

in other words, we have it coming first with the unity coefficient.

In order to get a unity coefficient, it's clear that I'm going to have to divide

every term in the equation above by l/Rth.

Or in other words, multiply every term by Rth over l.

So when I do that,

I will have di/dt +

Rth/l i = Rth/l isc.

Okay, there's our differential equation in standard form.