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

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2 MA 36600 LECTURE NOTES: MONDAY, MARCH 2 It makes sense to choose A and B such that 10 A + 2 B = 8 2 A + 10 B = 0 We can solve for A and B by multiplying the first equation by 5; or multiplying the second by 5: 50 A + 10 B = 40 2 A + 10 B = 0 52 A = 40 10 A + 2 B = 8 10 A + 50 B = 0 52 B = 8 This gives A = 40 / 52 = 10 / 13 and B = 8 / 52 = 2 / 13, so that one solution to the di ff erential equation is Y ( t ) = 10 13 e t cos 2 t + 2 13 e t sin 2 t. Hence the general solution to the nonhomogeneous di ff erential equation y 3 y 4 y = 8 e t cos 2 t must be the function y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) + Y ( t ) = c 1 e t + c 2 e 4 t + 10 13 e t cos 2 t + 2 13 e t sin 2 t for some constants c 1 and c 2 . Example 2. Finally, consider the di ff erential equation y 3 y 4 y = 2 e t . Again, we find the general solution. It su ces to find one solution Y = Y ( t ). If we attempt to make a guess in the form Y ( t ) = A e α t , we have Y 3 Y 4 Y = α 2 3 α 4 A e α t = 2 e t .
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