1. ## differential eq problem

A radar controlled tracking system is described by the following diff eqn:

X''(o)+100X'(o)+10,000X(o)= 10,000X(d)

X(d)=angular position of the target (in radians)
X(o)=angular position of the tracker (in rads)

If the target is moving at a rate of 0.5rad/s i.e. X(d) = 0.5t rads
what will the steady state error E be, where E=X(d) - X(o)

2. First, we split it into two equations:

$X''(o)+100X'(o)+10,000X(o)= 10,000X(d)$

Now, I'm not sure if you're familiar with D notation, but:

$D = \frac{d}{dx}$

$Dy = \frac{dy}{dx}$

Basically:

$D^n = \frac{d^n}{dx^n}$

So therefore:

$(D^2 + 100D + 10,000)X(o) = 0$

And the second equation:

$(D^2 + 100D + 10,000)X(o) = 10,000X(d)$

The Initial Steps

1. $D(e^{kx}) = ke^{kx}$ and $D^n(e^{kx}) = k^ne^{kx}$

2. Hence $(D^2 + 100D + 10,000)e^{\alpha x} = \alpha^2e^{\alpha x} + 100\alpha e^{\alpha x} + 10,000e^{\alpha x}$

$= (\alpha^2 + 100\alpha + 10,000)e^{\alpha x}$

$= \left(\alpha + \left(-50 + \frac{173.21i}{2}\right)\right)\left(\alpha + \left(-50 - \frac{173.21i}{2}\right)\right)e^{\alpha x}$

So the roots of the polynomial $\alpha^2 + 100\alpha + 10000$ are $50 \pm \frac{173.21i}{2}$

Hence $(D^2 + 100D + 10000)e^{\left(50 + \frac{173.21i}{2}\right)x} = 0$ and $(D^2 + 100D + 10000)e^{\left(50 - \frac{173.21i}{2}\right)x} = 0$

3. From 1 and 2 we deduce that for any quadratic function of D with constant coefficients that: $(aD^2 + bD + c) = e^{kx}(ak^2 + bk + c)$. This equals 0 if k is a root of $ak^2 + bk + c$.

First we deal with the Homogeneous Equation, for the solution we expect:

1. Two linearly independent solutions, $X_1(o)$ and $X_2(o)$.
2. Two arbitrary constants $c_1$ and $c_2$.
3. The general solution will be $X_c(o) = c_1y_1 + c_2y_2$.

First, we see that a possible solution is $X(o) = Ae^{kx}$

From the initial work, we can see that both roots are in fact complex and in the form $k = \alpha + i\beta$ or $k = \alpha - i\beta$, therefore, the general solution is of the form:

$X(o) = e^{\alpha x}[Pcos(\beta x) + Qsin(\beta x)]$

We have the roots:

$k_1 = 50 + \frac{173.21}{2}i$

$k_2 = 50 - \frac{173.21}{2}i$

So, therefore we arrive at the solution:

$X_c(o) = e^{50x}\left[Pcos\left(\frac{173.21}{2}x\right) + Qsin\left(\frac{173.21}{2}x\right)\right]$

Where:

$P = A + B$

and

$Q = i(A - B)$

Where A and B are constants.

Finally:

$X_c(o) = e^{50x}\left[(A + B)cos\left(\frac{173.21}{2}x\right) + i(A - B)sin\left(\frac{173.21}{2}x\right)\right]$

Nonhomogeneous Equation

The general solution of this is:

$y(x) = X_c(o) + X_p(o)$

Where $X_c(o)$ is the general equation we had above and $X_p(o)$ is a particular solution of the differential equation.

Ok, to solve this we need to go back to the formula:

$(D^2 + 100D + 10000)X(o) = 10000X(d)$

Got to go to class. If anyone else would like to solve it they can.