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lecture_19 (dragged) 1 - 2 MA 36600 LECTURE NOTES: MONDAY,...

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Unformatted text preview: 2 MA 36600 LECTURE NOTES: MONDAY, MARCH 2 It makes sense to choose A and B such that −10 A 2A + −2 B + −10 B = −8 = 0 We can solve for A and B by multiplying the first equation by −5; or multiplying the second by 5: 50 A 2A 52 A + 10 B + −10 B = 40 =0 = 40 −10 A + −2 B 10 A + −50 B −52 B = −8 = 0 = −8 This gives A = 40/52 = 10/13 and B = 8/52 = 2/13, so that one solution to the differential equation is 10 t 2t Y (t) = e cos 2t + e sin 2t. 13 13 Hence the general solution to the nonhomogeneous differential equation y ￿￿ − 3 y ￿ − 4 y = −8 et cos 2t must be the function y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) = c1 e −t + c2 e + 4t for some constants c1 and c2 . ￿ 10 t 2t e cos 2t + e sin 2t 13 13 ￿ Example 2. Finally, consider the differential equation y ￿￿ − 3 y ￿ − 4 y = 2 e−t . Again, we find the general solution. It suffices to find one solution Y = Y (t). If we attempt to make a guess in the form Y (t) = A eαt , we have ￿ ￿ Y ￿￿ − 3 Y ￿ − 4 Y = α2 − 3 α − 4 A eαt = 2 e−t . ￿ ￿ Upon choosing α = −1 would find that α2 − 3 α − 4 A eαt = 0 for all t. Hence a guess in this form would not work! This fails because the function f (t) = 2 e−t involves a polynomial of degree d = 0, and we must guess a solution which involves a polynomial of degree (d + 2) = 2. Make the guess that the solution is in the form ￿ ￿ Y (t) = A0 + A1 t + A2 t2 eαt where α = −1; and the Aj are coefficients yet to be determined. We have the derivatives Y Y￿ Y ￿￿ = A0 e−t = (−A0 + A1 ) e−t = (A0 − 2 A1 + 2 A2 ) e−t This gives the expression + A1 t e−t + (−A1 + 2 A2 ) t e−t + (A1 − 4 A2 ) t e−t + A2 t2 e−t + (−A2 ) t2 e−t + A2 t2 e−t ￿ ￿ ￿ ￿ ￿￿ −t −t Y − 3 Y − 4 Y = −5 A1 + 2 A2 e + −10 A2 t e + 0 t2 e−t = 2 e−t . ￿￿ ￿ It makes sense to choose A0 , A1 , and A2 such that −5 A1 + 2 A2 −10 A2 =2 =0 This gives A1 = −2/5 and A2 = 0. Since A0 does not appear in these equations, there is no harm in choosing A0 = 0, so that one solution to the differential equation is 2 Y = − t e−t . 5 Hence the general solution to the nonhomogeneous differential equation must be the function y ￿￿ − 3 y ￿ − 4 y = 2 e−t y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) = c1 e −t ￿ 2 + c2 e + − t e−t 5 4t ￿ ...
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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.

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