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Appendix: Differential Equations

 

You learned a bit about differential equations in first year, and you will learn more in second years maths. You will only need to solve some basic differential equations in this course.

Here is a summary of some basics.

First Order Differential Equation:


\begin{displaymath}\frac{dx}{dt} + a_0 x = f(t),
\ \ \ x(0)=x_0
\end{displaymath}

Second Order Differential Equation:


\begin{displaymath}\frac{d^2 x}{dt^2} + a_1 \frac{dx}{dt} + a_0 x = f(t),
\ \ \ x(0)=x_0, \frac{dx}{dt}(0)=\dot{x}_0
\end{displaymath}

General Solution:


x(t) = xh(t) + xp(t)

xh = complementary (homogeneous) solution, i.e. the solution of the homogeneous equation (forcing term f=0):

\begin{displaymath}\frac{d^2 x}{dt^2} + a_1 \frac{dx}{dt} + a_0 x = 0
\end{displaymath}

xp= particular solution, the part that is determined by the forcing term f.

Homogeneous Solution: Trial solution

x(t) = Kest

Differentiate and plug into homogeneous equation gives the characteristic equation


s2 + a1 s + a0 = 0

Two solutions


\begin{displaymath}s_{1,2} = \frac{-a_1 \pm \sqrt{a_1^2 - 4a_0}}{2}
\end{displaymath}

So homogeneous solution:

xh(t) = K1 es1 t + K2 es2 t

Re-write characteristic equation

\begin{displaymath}s^2 + 2 \zeta \omega_0 s + \omega_0^2 = 0
\end{displaymath}

$\zeta = \frac{a_1}{2\sqrt{a_0}} = $ damping ratio

$\omega_0 = + \sqrt{a_0} = $ unforced natural frequency

Write

\begin{displaymath}s = \sigma + j \omega
\end{displaymath}

Three cases:

Overdamped
$\zeta > 1$ Two distinct real roots

\begin{displaymath}s_1 = \sigma_1, \ \ \ s_2 = \sigma_2
\end{displaymath}

Homogeneous solution

\begin{displaymath}x_n(t) = K_1 e^{\sigma_1 t} + K_2 e^{\sigma_2 t}
\end{displaymath}

Critical damping
$\zeta = 1$ Two equal real roots

\begin{displaymath}s_1, s_2 = \sigma
\end{displaymath}

Homogeneous solution

\begin{displaymath}x_h(t) = K_1 e^{\sigma t} + K_2 t e^{\sigma t}
\end{displaymath}

Underdamped
$\zeta < 1$ Two distinct complex roots

\begin{displaymath}s_1 = \sigma + j \omega, \ \ \ s_2 = \sigma - j \omega
\end{displaymath}

Homogeneous solution

\begin{displaymath}x_n(t) = K_1 e^{\sigma t} \cos ( \omega t) + K_2 e^{\sigma t} \sin (\omega t)
\end{displaymath}

Particular Solution: The trial form of the particular solution xp(t) depends on the forcing function f(t).

General Solution: the sum of the homogeneous and the particular solutions

x(t) = xh(t) + xp(t) = K1 es1 t + K2 es2 t + xp(t)

Method of Solution:

To solve a linear differential equation with given intital condition, the general method of solution is:

1.
Write down the characteristic equation and solve for the two zeros s1 and s2.
2.
Determine the form of the homogeneous solution (Keat, Kteat, $K_1e^{\sigma t} \cos( \omega t) + K_2e^{\sigma t}\sin (\omega t)$, etc) and write it down.
3.
Guess the form of the particular solution xp(t), based on the form of f(t). Substitute xp(t) into the DE and find any constants (A< B, etc) in your formula for xp(t).

4.
Add the particular solution to the homogeneous solution:

x(t)= xh(t) + xp(t)

and now use the initial conditions to determine the constants (K1, K2, etc) in the expression for the homogeneous xh(t). This can only be done after any constants in the particular soilution have been determined (in the previous step).

5.
You now have the complete solution. As a check, plug it into the DE to see if it is correct and also you can check the initial conditions.
6.
Also, when solving circuit problems, is your solution consistent with your intuition about how the waveform should look? Does your solution agree with a spice or maple simulation?


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