Hint as to how to solve an integral differential equation

[barryj]

I know how to solve a differential equation using Eulers method but what if the equation has an integral part?

i.e. a RLC electrical circuit.

Vsource = iR + L di/dt + (1/C)int i dt

can this be done? a link to how to solve this would be helpful.

 

[scottdave]

One way is to turn it into a 2nd order diffeq. Since electric current is the time derivative of change in charge (Q), you have I = dQ/dt and di/dt = d²Q / dt². Another way is to use Laplace Transforms, if you've learned that.

 

[barryj]

I am reading a book called "The Innovators". It mentions a difference engine and a differential analyzer, that is digital, that solves differential equations.
I assume that the differential/integral equation is transformed into a difference equation that is solved numerically. I want to know how this is done. It seems like one would assume a small value for deltaT and then incrementally solve the problems.

 

[barryj]

scottdave. can a 2nd order diff eq be solved numericlly?

 

[scottdave]

scottdave. can a 2nd order diff eq be solved numericlly?

Yes. You change it into 2 first order equations which are related. I'm posting a couple of links.
https://math.oregonstate.edu/home/p...uestStudyGuides/ode/second/so_num/so_num.html

And there's this YouTube video which I found. I haven't watched the whole thing yet, but the first part looks pretty good.

 

[barryj]

Thanks, I think I have enough to go forward.

 

[AVBs2Systems]

Hello.

The differential equation for a second order system like an RLC circuit (series) driven by a voltage source and an RLC circuit (parallel) driven by a current source is derived, as suggested above, by taking the derivative on both sides:

$$
v(t) = i(t) R + L i'(t) + \dfrac{1}{C} \displaystyle \int i(t) \,\,\,\, \text{dt}
$$
You take the first derivative, on both sides, and multiply by one over L
$$
\dfrac{v'(t)}{L} = i''(t) + i'(t) \dfrac{ R}{L} + \dfrac{ i(t)}{LC}
$$
Now we redefine the constants:
$$
\text{Damping ratio} \,\,\,\,\,\delta = \dfrac{R}{2L} \,\,\,\,\, \text{Resonant frequency squared} \,\,\,\,\, {\omega_{r}}^{2} = \dfrac{1}{LC}
$$
Then the equation becomes:
$$
\dfrac{v'(t)}{L} = i''(t) + 2 \delta i'(t) + {\omega_{r}}^{2} i(t)
$$
The same derivation is done for a parallel RLC circuit where voltage is the same (as they are in parallel):
$$
i(t) = \dfrac{v(t)}{R} + \dfrac{1}{L} \displaystyle \int v(t) \,\,\,\,\, \text{dt} + C v'(t)
$$
Multiplying throughout by one over C, after taking the first derivative:
$$
\dfrac{i'(t)}{C} = v''(t) + \dfrac{v'(t)}{RC} + \dfrac{v(t)}{LC}
$$
You redefine the constants:
$$
\text{Damping ratio} \,\,\,\,\,\alpha = \dfrac{1}{2RC} \,\,\,\,\, \text{Resonant frequency squared} \,\,\,\,\, {\omega_{r}}^{2} = \dfrac{1}{LC}
$$
The equation then becomes similar to the series one:
$$
\dfrac{i'(t)}{C} = v''(t) + 2 \alpha {v'(t)}+ {\omega_{r}}^{2} {v(t)}
$$
The characteristic equations of both the differential equations are the same:
$$
x^{2} + 2 cx + {\omega_{r}}^{2}
$$
Where $\text{DF = the damping factor for a parallel or series configuration as defined earlier}$
You solve for the homogeneous solution using the quadratic equation (as is used for second order LINEAR ODEs):
$$
x ={ - \text{DF}\pm \dfrac{\sqrt{4(\text{DF}^{2} - {\omega_{r}}^{2}) } }{2} }{ }
$$
The homogenous response depends on the values of the damping factor and resonant frequency. The particular solution of this differential equation depends on the nature of the forcing function, exponential forcing functions result in exponential, purely sinusoidal can be solved using impedences, others with their corresponding methods.

$$
\textbf{The laplace transform: Finding the homogeneous solution, particular, and with initial values (if given). }
$$
The laplace transform is thus preferred to solve such linear ODEs with constant coefficients.
The transform allows you to express the output, current or voltage as well as the homogeneous (source free) response, in one go, as well as considering the initial conditions of the circuit. So its probably the easiest and least time consuming method for solving such ODEs.