First order circuits. Solving for v(t)

[CoolDude420]

Homework Statement

So our lecture introduced first order circuits to us.
We are trying to solve for v(t) in the following circuit. However in the notes he gets it into a form in which we can solve for v(t). However I can't seem to get it into the same form he has. This is more of a maths question rather than a circuits question.

The purpose of writing it in the form that he has is to solve for v(t). Any ideas how I can get it into that form?

Homework Equations

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i(t) = C dv(t)/dt

The Attempt at a Solution

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In Pic above

 

[NascentOxygen]

Presumably E is a constant here?

The derivative of a function is unaffected if you first add (or subtract) a constant value^∆[/size] to the function,
so it follows that d(v(t))/dt is identical to d(v(t)–E)/dt

It looks like your lecturer might be finding it convenient to use this fact in his derivation of v(t)?

^∆ the derivative of a sum is equal to the sum of the individual derivatives

 

[CoolDude420]

Presumably E is a constant here?

The derivative of a function is unaffected if you first add (or subtract) a constant value^∆ to the function,
so it follows that d(v(t))/dt is identical to d(v(t)–E)/dt

It looks like your lecturer might be finding it convenient to use this fact in his derivation of v(t)?

^∆ the derivative of a sum is equal to the sum of the individual derivatives

Ah. That makes much more sense. I've been trying to figure this out for a while now. Thank you very much,

 

[CoolDude420]

Hi,

I've just come back to revise this after properly I think understanding dynamic circuits. I am told that in the DC steady state(i.e when the source is providing a steady voltage), capacitors behave as open circuits and inductors behave as short circuits. So why doesn't that happen in this case(where E is constant).? I mean the circuit is providing a constant V and so the time derivative in i = C dv/dt should evaluate to 0 thus giving us i=0 which is an open circuit. Why don't we just replace the capacitor with a open circuit here?

 

[NascentOxygen]

Why don't we just replace the capacitor with a open circuit here?

You can. If you do that it will tell you that the steady-state current is zero, which is true.