Question

Can someone help me with this problem?. Ri(t)+Ldi(t)dt+1/C∫i(t)dt=E(t)

Answer 1

(1)
I have this equation (RLC circuit):
\[Ri(t)+L \frac{ di(t) }{ dt }+\frac{ 1 }{ C }\int\limits_{?}^{?}i(t)dt=E(t)\]
We use this:
\[x_{1}(t)=\int\limits_{?}^{?}i(t)dt\]
\[x_{2}=\frac{ dx_1{1}(t) }{ dt }=i(t)\]
(2)
Then we substitute:
\[R \frac{ dx_{1}(t) }{ dt }+L \frac{ d^{2}x_{1}(t) }{ dt^{2} }+\frac{ 1 }{ C }x_{1}(t)=E(t)\]
And :
\[x_{2}=\frac{ dx_1{1}(t) }{ dt }=i(t)\], so
\[\frac{ di(t) }{ dt }=\frac{ d^{2}x_{1}(t) }{ dt^{2} }=\frac{ dx_{2}(t) }{ dt }\]
.My question is how can I decompose the (2) in differential equations of first order?. Because I didn't see that in the last semester. Can you explain me with a different example, because I want to do this by my own. Thanks
One of them is:
\[\frac{ dx_{1} }{ dt }=x_{2}(t)\]. But I don't know why

Answer 2

I dont really understand what is that you are asking for.
In this case \(x_1(t)\) is the charge Q stored in time interval \(t_1-t_0\).
So dQ/dt=i(t)

Answer 3

This is a sentences:
A differential equation of n-th order can be decomposed into n-th differential equations of first order. But I don't know how

Answer 4

ok. I will try to explain...

Answer 5

imagine you have 2º order ODE:
y''=f(x,y,y')
then you can introduce 2 new variables:
u=y
and
v=y'
Using this, we see that u'=v, right till here?

Answer 6

Yes

Answer 7

so, continuing v'=y''=f(x,y,y') your original equation.
So you got two equations:
1º: u'=v
2º v'=f(x,u,v)
which represent same original relation

Answer 8

Ahh, ok.

Answer 9

I think I got it thanks.

Answer 10

How it is called this method in English?