User:Egm6321.f10.team4.Yoon/Mtg10

 Mtg 10: Tue, 14 Sep 10

[[media: 2010_09_14_15_00_52.djvu | Page 10-1]]
On the lecture note [[media: 2010_09_09_14_48_41.djvu | (2)p.9-4 ]] we have,

Apply 2nd exactness condition to find h:

For (1) to be exact:

Solving (5) for integrating factor h(x,y) is usually not easy. So, we will consider 2 particular cases:

case 1: Suppose $$\displaystyle h_y = 0 $$; then (5) becomes:

[[media: 2010_09_14_15_00_52.djvu | Page 10-2]]
case 2: Assume $$\displaystyle h_x=0 \Rightarrow h $$ is a function of y only.

Read F09 lecture notes and find the expression for for h(y) for case2

Application: General non-homogenous L1_ODE_VC

If $$\displaystyle P(x) \neq 0; \ \ \ \ \ \forall x $$

[[media: 2010_09_14_15_00_52.djvu | Page 10-3]]
Eqn.(1) [[media: 2010_09_14_15_00_52.djvu | p.10-2:]]

Eqn.(5) [[media: 2010_09_14_15_00_52.djvu | p.10-2:]]

Recall,

Application: Non-homog L1_ODE_VC

Read F09 $$\displaystyle h(x)=x$$ and thus $$\displaystyle y(x) = \frac{x^3}{4}+\frac{k}{x}$$