09 review exam 1, linear vs nonlinear

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

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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 cant be extended past x 1 . Here are two examples to illustrate this behavior. (Theyre not so very important for the review, so you should skip them if youre 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 cant 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 its defined near x = 0, as the Existence Theorem requires); but it goes to infinity as x approaches 1, so it cant be extended beyond that point. Next thing you should know is Eulers 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 Eulers method gives y ( x + h ) y + hf ( x ,y ) = y 1 ....
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09 review exam 1, linear vs nonlinear - 18.03 Lecture #9...

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