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Unformatted text preview: 18.03 Lecture #4 Sept. 16, 2009: notes Two topics for today. First (thanks to Professor David Jerison for explaining it to me) well look at a way to think about the solution method for linear ODE. One of the points of this course is to learn how to make solutions do what you want them to do, and this is supposed to help. Second, well look at a few other other special classes of differential equations for which one can find formulas for solutions. For topic one, well think about Money. The Big Idea is that in a linear ODE, you can think of the coefficient P ( x ) of y as like (the negative of) a variable interest rate, and the function Q ( x ) as a variable rate of deposit into the account. The formula (Growth with Deposits) below says that each bit of deposit grows in some exponential way; adding up (integrating) all those bits gives the solution to the equation that we did yesterday. Suppose your bank pays interest on a saving balance at some rate I ( x ) that may change with time. (Since time is measured in years, I ( x ) might vary over some range like . 01 to . 10.) Roughly speaking, this means that each year the bank adds I ( x ) b ( x ) to your balance b(x). (This is annual componding.) A little more precisely, it means that each month the bank adds I ( x ) b ( x ) / 12 to your balance. (This is monthly compounding. Even more precisely, it means that each day the bank adds I ( x ) b ( x ) / 365 to your balance. Really and truly, it means that the bank solves the differential equation db dx = I ( x ) b ( x ) (Interest) to keep track of your balance. If the interest rate is a constant I , then your balance grows expo- nentially: b ( x ) = Ae Ix . (Growth with Constant Interest) If the interest rate is not constant, then you can solve (Interest) by separation of variables, leading...
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- Fall '09