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# AllLectures - LECTURE NOTES Math 16C Short Calculus Spring...

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Unformatted text preview: LECTURE NOTES Math 16C Short Calculus Spring 2008, Section 2 Dr. Peter Malkin May 27, 2008 1 Section C.1: Solutions of differential equations Points to cover (Section C.1): • What is a differential equation? • What is a general solution of a D.E.? • What is a particular solution of a D.E.? Definition 1 A differential equation is an equation involving a differentiable function (e.g. y = f ( x ) ) and one or more of its derivatives (e.g. y ′ = dy dx and y ′′ = d 2 y dx 2 ). E.g. y ′′ − 4 y ′ + 3 y = 0 or equivalently d 2 y dx 2 − 4 dy dx + 3 y = 0 ( ∗ ). Definition 2 A solution of a D.E. is a function y = f ( x ) that satisfies it. E.g. y = 4 e x + 5 e 3 x is a solution of ( ∗ ). Question: Verify that y = 4 e x + 5 e 3 x is a solution of ( ∗ ). Answer: y ′ = dy dx = d dx bracketleftbig 4 e x + 5 e 3 x bracketrightbig = 4 d dx [ e x ] + 5 d dx bracketleftbig e 3 x bracketrightbig = 4 e x + 15 e 3 x. y ′′ = d 2 y dx 2 = d dx bracketleftbigg dy dx bracketrightbigg = d dx bracketleftbig 4 e x + 15 e 3 x bracketrightbig = 4 e x + 45 e 3 x . Then, putting y , y ′ and y ′′ into ( ∗ ), we have y ′′ − 4 y ′ + 3 y = bracketleftbig 4 e x + 45 e 3 x bracketrightbig − 4 bracketleftbig 4 e x + 15 e 3 x bracketrightbig + 3 bracketleftbig 4 e x + 5 e 3 x bracketrightbig = 4 e x + 45 e 3 x − 16 e x − 60 e 3 x + 13 e x + 15 e 3 x = [4 − 16 + 12] e x + [45 − 60 + 15] e 3 x = 0 . Is this the only solution of ( ∗ )? NO! The function y = − 2 e x + 11 e 3 x is also a solution of ( ∗ ). Check this yourselves. In fact, there are infinitely many different solutions of ( ∗ )! Consider y = Ae x + Be 3 x where A and B are constants. Then, dy dx = Ae x + 3 Be 3 x and d 2 y dx 2 = Ae x + 9 Be 3 x . So, y ′′ − 4 y ′ + 3 y = bracketleftbig Ae x + 9 Be 3 x bracketrightbig − 4 bracketleftbig Ae x + 3 Be 3 x bracketrightbig + 3 bracketleftbig 4 e x + 5 e 3 x bracketrightbig = Ae x + 9 Be 3 x − 4 Ae x − 12 Be 3 x + 3 Ae x + 3 Be 3 x = [ A − 4 A + 3 A ] e x + [9 B − 12 B + 3 B ] e 3 x = 0 . There is one solution of ( ∗ ) for every choice of A and B . The function y = Ae x + Be 3 x is called a general solution of the D.E. A particular solution is a solution where A and B are fixed. • A general solution of a D.E. has unspecified constants. • A particular solution of a D.E. has no unspecified constants. E.g. y = 4 e x + 5 e 3 x and y = − 2 e x + 11 e 3 x are particular solutions of ( ∗ ). 2 The particular solutions of a D.E. are obtained from the general solution and some initial condi- tions on the general solution and its derivatives. We use the initial conditions to find the values of the constants in the general solution....
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## This note was uploaded on 06/15/2011 for the course CAL 3 taught by Professor Smith during the Spring '11 term at Arkansas State.

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AllLectures - LECTURE NOTES Math 16C Short Calculus Spring...

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