I am trying to find v(t) at T>0.

My Attempt
KVL(right loop): 5v = C*10000 \(\frac{dv}{dt}\) + v

Solving this gives \(\frac{dv}{dt}\) + \(\frac{v}{C*10000}\) = \(\frac{1}{C*2000}\)

From here Tau = \(\frac{1}{(1*10^{-6})*10000}\)

v(t) = \(5e^{-.01t}\)

 

In addition to the abysmal failure to properly track units (which is almost certainly because you were never expected to do so), or even bother to tack units onto the final result (which you probably are at least expected to do), you failed to ask if the answer makes sense.

At t = 0 s, what is the voltage across the capacitor?

What does your result say the voltage across the capacitor is at t = 0 s.

 

Is the object here to solve the differential equation, or find the voltage across the capacitor? If the latter, then look at the capacitor in the circuit.

Does the capacitor start with an initial charge? What is the capacitor voltage at t=(0-), and at t=(0+)? Unless there is an impulse function at t=0, will the capacitor voltage change at t=0? Is the circuit charging or discharging the capacitor? What is the initial capacitor voltage? What is the final capacitor voltage? Will v(t) follow an exponential rise or an exponential decay from the initial to the final value? This should all be evident by inspection. Sketch the behavior of v(t) over six time-constants. From the sketch of v(t) you should be able to write the equation for v(t) directly.

 

I suspect, at this point in many first circuits courses, that they are just starting with transient response in first order circuits, so they may not yet have the background to do everything that you suggested. But they should be able to do some of it. Being able to develop and solve the differential equation is intended to help bridge the gap between the two.

 

In addition to the abysmal failure to properly track units (which is almost certainly because you were never expected to do so), or even bother to tack units onto the final result (which you probably are at least expected to do), you failed to ask if the answer makes sense.

At t = 0 s, what is the voltage across the capacitor?

What does your result say the voltage across the capacitor is at t = 0 s.

At t=0 my equation suggests that v=5v. This is not true as it should be 0 because capacitors cannot change their voltage instantaneous.

So would I just need to add to the equation to make v=0v when t=0?

\(v(t) = 5(1-e^{-100t})\) This equation makes it so that for t=0 the result would be 5(1-1) = 0v.

Edit: Tau in OP should be 100 also.

 

I suspect, at this point in many first circuits courses, that they are just starting with transient response in first order circuits,

The only educational value to be derived from transient analysis by differential equation is to convince the student that first-order RC or RL circuits excited by step functions (ON-OFF switches), which are the most common situation encountered, always exhibit an exponential rise or exponential decay response that can be completely characterized by an initial value, final value, and time constant.

 

At t=0 my equation suggests that v=5v. This is not true as it should be 0 because capacitors cannot change their voltage instantaneous.

Correct. These are the kinds of questions you want to get in the routine habit of asking yourself both as you are solving problems and once you get to the final result.

So would I just need to add to the equation to make v=0v when t=0?

\(v(t) = 5(1-e^{-100t})\) This equation makes it so that for t=0 the result would be 5(1-1) = 0v.

This is a very dangerous approach -- and I suspect it is how you got the wrong equation in the first place. By "just need to add to the equation" you are practicing the Art of the Happening, which comes down to throwing things around more-or-less randomly and hoping that, eventually, you'll stumble across something that happens to work.

Edit: Tau in OP should be 100 also.

And this error didn't get caught because you didn't track your units. If you had, you would have seen that your exponent ended up with units of time-squared, and since the exponent must be dimensionless, you would have known that the answer had to be wrong.

 

The only educational value to be derived from transient analysis by differential equation is to convince the student that first-order RC or RL circuits excited by step functions (ON-OFF switches), which are the most common situation encountered, always exhibit an exponential rise or exponential decay response that can be completely characterized by an initial value, final value, and time constant.

Your point?

You've just described the educational value it has. So it is not surprising that introductory circuits courses spend a little bit of time having students analyze the transient response of first (and second) order circuits via solving the governing differential equations specifically to demonstrate and convince them of the reasonableness and validity of writing down the solutions by inspection or through much more efficient analysis techniques aimed at identifying the relevant parameters, such as the initial value, final value, and time constant.