14 particular solns, exp response formula

# 14 particular solns, exp response formula - 18.03...

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Unformatted text preview: 18.03 Lecture #14 Oct. 9, 2009: notes The idea today is to work with formal properties of exponential functions from calculus, espe- cially with the formula d ( e rx ) dx = re rx , (Exponential derivative) to find explicit solutions to linear ODE with constant coefficients. I’ll first (re)derive the facts I’ve already given at least for second order equations, relating the roots of the characteristic polyno- mial to solutions of the homogeneous equation. Then I’ll look at finding particular solutions to inhomogeneous (still constant coefficient) equations. Definition. The differential equation [ D n + a n- 1 D n- 1 + ··· + a 1 D + a ] y = b ( x ) (Constant coefficient n th order) is called a constant coefficient (because all the coefficients a i are constant functions of x ) linear differential equation . The polynomial function p ( q ) = q n + a n- 1 q n- 1 + ··· + a 1 q + a is called the characteristic polynomial of the differential equation. In terms of p , we can write the differential equation as p ( D ) y = b ( x ). The corresponding homogeneous equation is [ D n + a n- 1 D n- 1 + ··· + a 1 D + a ] y = 0 , (Homogeneous constant coefficient n th order) or just p ( D ) y = 0. I pointed out in Lecture 11 that D ( e rx ) = re rx . An immediate consequence is Theorem: Substitution Rule Suppose p is a polynomial. Then for any complex number r , p ( D ) e rx = p ( r ) e rx . In particular, the exponential function e rx is a solution of the (homogenous constant coefficient linear) ODE p ( D ) y = 0 if and only if p ( r ) = 0 . The reason for the name is this: to apply p ( D ) to e rx , you substitute the number r for D . Proof. Write p ( q ) = a n q n + a n- 1 q n- 1 + ··· + a 1 q + a . By repeated application of (Exponential derivative), we get D m e rx = r m e rx . Multiplying by the constant a m gives a m D m e rx = a m r m e rx . Now sum over m and factor out the common e rx : [ a n D n + a n- 1 D n- 1 + ··· + a 1 D + a ] e rx = [ a n r n + a n- 1 r n- 1 + ··· + a 1 r + a ] e rx . The term in brackets on the left is by definition p ( D ), and the term in brackets on the right is p ( r ). So this is the formula we wanted to prove. Q.E.D. 1 Recall that one of the reasons complex numbers are so useful is Fundamental Theorem of Algebra. Suppose p ( q ) = q n + a n- 1 q n- 1 + ··· + a 1 q + a is a polynomial of degree n with coefficients a i in C . Then p can be factored into linear factors: there are n complex numbers r 1 ,... ,r n so that p (...
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14 particular solns, exp response formula - 18.03...

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