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Unformatted text preview: 2 MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23 Upon multiplying the first equation by g ￿ (t) and the second by −g (t), we find the following system of equations: f (t) g ￿ (t) c1 + g (t) g ￿ (t) c2 = 0 ￿ −f (t) g (t) c1 + −g (t) g ￿ (t) c2 = 0 ￿ ￿ f (t) g ￿ (t) − f ￿ (t) g (t) c1 =0 We conclude that ￿￿ ￿￿ W f, g (t) · c1 = W f, g (t) · c2 = 0 for all t. ￿￿ Since not both c1 and c2 are zero, this implies W f, g (t) = 0 for all t. Clearly this is a contradiction, so f and g must be linearly independent. This shows p2 =⇒ p3 . Now we show that p3 =⇒ p1 , so ￿ ￿ that assume ￿￿ f and g are linearly independent. If there did not exist t0 such that W f, g (t0 ) ￿= 0, then W f, g (t) = 0 for all t. In particular, the Quotient Rule for Derivatives gives the expression ￿￿ ￿ ￿ W f, g (t) d g (t) f (t) g ￿ (t) − f ￿ (t) g (t) g (t) = = =0 =⇒ =λ 2 2 dt f (t) f (t) f (t) f (t) must ￿ a￿constant. Hence g = λ f so that f and g are linearly dependent. Clearly this is a contradiction, be so W f, g (t0 ) ￿= 0 for some t0 . This shows p3 =⇒ p1 . We have shown that p1 =⇒ p2 and p2 =⇒ p3 and p3 =⇒ p1 . The three statements p1 =⇒ p3 and p3 =⇒ p2 and p2 =⇒ p1 follow from the compositions p1 =⇒ p2 =⇒ p3 p3 =⇒ p1 =⇒ p2 p2 =⇒ p3 =⇒ p1 This shows that pi =⇒ pj holds for all i and j . This completes the proof. Example. Consider the functions f (t) = et and g (t) = e2t . We will show that f and g are linearly independent using two different methods. First we show that they are independent using the original definition. Consider the equation c1 f (t) + c2 g (t) = 0 for all t. The function y (t) = c1 f (t) + c2 g (t) must be the zero function, so in particular its derivative y ￿ (t) = c1 f ￿ (t) + c2 g ￿ (t) is also the zero function. Hence we have the system of equations c1 et c1 et + c2 e2t + 2 c2 e2t =0 =0 =⇒ c1 c1 + c2 + 2 c2 =0 =0 Upon subtracting the first equation from the second, we see that c1 = c2 = 0. Since this is the only solution to the equation c1 f + c2 g = 0, we see that f and g must be linearly independent. Second we show that they are independent using the Wronskian. The Wronskian of f and g is the function ￿￿ ￿ ￿￿ ￿ ￿ ￿￿ ￿ W f, g (t) = f g ￿ − f ￿ g = et 2 e2t − et e2t = 2 e3t − e3t = e3t ￿= 0. This this function is nonzero for all t, we see that f and g must be linearly independent. Fundamental Solutions Revisited. We return to the homogeneous equation a(t) y ￿￿ + b(t) y ￿ + c(t) y = 0. Say that we have two solutions y1 = y1 (t) and y2 = y2 (t). When is the linear combination y (t) = c1 y1 (t) + c2 y2 (t) the general solution to the differential equation? If y1 and y2 are linearly independent, then y = c1 y1 + c2 y2 is the general solution. We prove this as follows. Say that we have the initial conditions y (t0 ) = y0 and ￿ y ￿ (t0 ) = y0 . We showed in the previous lecture that if the quantity ￿ ￿ W0 = y1 (t0 ) y2 (t0 ) − y1 (t0 ) y2 (t0 ) ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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