1
Linear, First Order, ODE: Integrating Factor
Consider the following linear dierential equation for y(t)
y + p(t)y = g(t)
(1)
where p(t) and g(t) are known functions of t.
To nd the analytical solution, rst nd the integrating factor (I.F.) given as (t),
12
w,,.. .. A 5. . ____._.
i :' " sagas Ana-54' __
atriwbc. mons accw in Engimenng i1
fa-zobeA'nSVThejr repmsenL-wbon uh W3 of Simple. 3
gpwwc funcbo-ns, each m sine, and cos-in: , a a. -
it-ear 0f gma-L' pmcH-ml importancb This leads 1
gift
.7 - . brookTaLJ-[W MEW
Viki-£0" 365-43 EanASIOn : Memdan)
Comic-Jar a f/anbon 35(90- Lek; assume that we
«new 'Ht Velma if Hm m¢Hon @ 1:61 (ix- f(a) a
KMwn} We; also assume Hm? «rst (n+1) clerWa-ths
of {(1) @ 1: are mdnuous an M inHJva
Whining 0
1
Explicit Euler Matlab Code for Shooting Method
Following code solves the dierential equation using Forward Euler discretization and also compares the numerical
solution to the exact solution.
d2 T
+ H(Ta T ) = 0;
dx2
1
2
T (0) = 40;
T (10) = 200
(1)
% E
Handout: Stability and Accuracy
1
Stability
Stability of a system is associated with the solution of the governing equations for specied initial or boundary
conditions. Here we are interested in systems where the exact solution (if known) is well-behaved
1
Taylor Series Expansion
Consider a continuous function f (x). Then, the Taylor Series expansion of the function around a point xj is:
f (x) = f (xj ) + (x xj )f (xj ) +
(x xj )2
(x xj )3
(x xj )4
f (xj ) +
f (xj ) +
f
2!
3!
4!
(xj ) + .
(1)
In the abov
1
Wronskian and linearly independent solution
For a second-order dierential equation for y(t), y1 and y2 are two linearly independent solutions
if the Wronskian (W) is non-zero. Here the Wronskian is dened as:
W =
y1 y2
y1 y2
(1)
Example: Consider
y + 5y
Handout: Root Finding
1
Newtons Method
Newtons method for solving f (x) = 0 draws the tangent to the graph of f (x) at any point and
determines where the tangent intersects the x axis. The method requires one starting value, x0
(a guessed value). The iter