09 review exam 1, linear vs nonlinear

# 09 review exam 1, linear vs nonlinear - 18.03 Lecture#9...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 18.03 Lecture #9 Sept. 28, 2009: notes ANNOUNCEMENT : the exam takes place this Wednesday during class time but NOT HERE ; instead it is on the third floor of Walker (old building between here and the river). No new material today; just a summary of all the old material, and how to put it all together to solve random ODE. A reasonable first step in dealing with a new differential equation is to write it in standard form dy dx = f ( x,y ) . (Standard ODE) Typically there will also be an initial condition y ( x ) = y . (Initial condition) For general equations like this (with f a continuous function) you should know the Existence The- orem , which says that there a function y satisfying both (Standard ODE) and (Initial Condition), and defined for x close to x . There are two reasons that the solution might not be defined for all values of x . First, when you try to extend y , the points ( x,y ( x )) on the graph can move outside of the domain where f is defined; so (Standard ODE) stops making sense. Second, the values of y can tend to ∞ as x approaches some finite value x 1 ; so the solution can’t be extended past x 1 . Here are two examples to illustrate this behavior. (They’re not so very important for the review, so you should skip them if you’re reading this on Wednesday afternoon.) Example: solution goes out of the domain of f . Consider the ODE y ′ = ln( x ) , y (1) = − 1 . The function f ( x,y ) = ln( x ) is defined only when x > 0, so it only makes sense to talk about a solution for x > 0. The solution is y ( x ) = x ln( x ) − x ( x > 0) . (by 18.01). Even though this function tends nicely to zero as x approaches zero (rather than blowing up), we can’t talk about a solution except for x > 0. Example: solution goes to ∞ . Consider the ODE y ′ = y 2 , y (0) = 1 . This differential equation is defined for all values of x and y , so nothing like the first example can go wrong. But the solution is y ( x ) = 1 1 − x . This solution is defined for −∞ < x < 1 (and in particular it’s defined near x = 0, as the Existence Theorem requires); but it goes to infinity as x approaches 1, so it can’t be extended beyond that point. Next thing you should know is Euler’s method to calculate approximate solutions numerically. Remember that you start by fixing a small “step size” h . (Smaller step sizes mean more calculation, but more accurate answers.) The idea is to use the linear approximation to y to estimate y ( x + h ) ≈ y ( x ) + hy ′ ( x ) . 1 We know the value y ( x ) = y from (Initial Condition), and we know the derivative y ′ ( x ) = f ( x ,y ) from (Standard ODE). So the first step in Euler’s method gives y ( x + h ) ≈ y + hf ( x ,y ) = y 1 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

09 review exam 1, linear vs nonlinear - 18.03 Lecture#9...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online