Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 2: Equations of Order One

Similar presentations


Presentation on theme: "Chapter 2: Equations of Order One"— Presentation transcript:

1 Chapter 2: Equations of Order One
MATH 374 Lecture 4 Chapter 2: Equations of Order One

2 2.1: Separation of Variables
The differential equation is said to have variables separable if it is of the form From (2) we can write which can be integrated to obtain a solution to (1)!! This technique is called separation of variables. Note that (2) is equivalent to the equation M(x) dx + N(y) dy = 0, with M(x) = g(x) and N(y) = -1/h(y). (See text, page 43.) 2

3 Example 1 Solve Solution: Note that since
are continuous for all (x,y), (4) will have a unique solution through any point (x0, y0)! With h(y) = 4y2 and g(x) = x, it is clear that we can use separation of variables! 3

4 Example 1 (continued) 4 Have we found all solutions to (4)?
Note that (0,0) must have a solution to (4) passing through it (why?), but when we put x = 0 and y = 0 into (5), we find: Clearly (6) has no solution for C1 in R! Looking at (4) again, we find that y  0 is also a solution to (4). Since putting any (x0, y0) with y0  0 into (5) will yield a C1, Theorem 1.1 implies that all solutions to (4) are given by: 4

5 Notes We call (5) a general solution of (4). We call (6) the general solution of (4). In Example 1, we “lost” the solution y ´ 0 in the separation of variables procedure, when we divided by h(y) = 4y2. In general, if separation of variables is used, we will lose those solutions which satisfy initial condition (x0, y0) with h(y0) = 0. To each root of h(y) there corresponds a solution to (2) of the form y ´ y0.

6 Example 2 Solve Solution:

7 Example 3 Find the particular solution to Solution:

8 Example 3 (continued)

9 Example 3 (continued)


Download ppt "Chapter 2: Equations of Order One"

Similar presentations


Ads by Google