MIT differential equation---7,8

First order ODE linear with constant coefficients

y' + ky = kqe(t)

First see another ODE:

y' + ky = q(t)

solution will be in the form of :

y = e^(-kt) $q(t)e^kt  dt  + ce^(-kt)

Now let's consider steady state solution again. As time goes to infinity, the unstable solutions will be asymptotic to the stable solution.

But what's the stable solution for the function above. There's certain vagueness. The general idea is to pick up the simplest. 

Input is the right hand side. Recall the water tank example. Input is q(t) , the temperature of the water bath

Response, or the output is the solution

Back to the original equation .  y' + ky = kqe(t)

(1/k) y' + y = qe(t)

Superposition principle:

q1(t) -> y1(t)

q2(t) ->y2(t)

q1(t) + q2(t) -> y1(t) + y2(t)

Notice: this corresponds to the linearity of ODE. It will be totally inapplicable to other forms.

Now, let's talk about the situation when the input is trigonometric. kq(t) = coswt

w is the so-called angular frequency, which means # of oscillation in 2PI.

We solve this equation by using the complex number. To use the complex number, we complexify the problem.

From real number domain go to complex number domain and  then go back. Easier to exponential. 

coswt = RE(e^iw). We will now call y y_tilt = y1 + i y2

Then, we can use integrating factor to solve the equation. The solution will be the real part of the Y.

Up to this point, we assume k > 0. So what happens when k < 0? Although the technique holds the same, the results of state, input, output don't hold.

ODE comes back to application. In reality, some k maybe not meaningful.

The simplest population model: dP/dt = aP.   [a] is positive here, so the population is growing.