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4 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 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 + 1 2 e 2 t for some constants c 1 and c 2 . Example. Consider the di ff erential equation y 3 y 4 y = 2 sin t. Again, we will find the general solution of this equation. To do so, we compute functions y 1 = y 1 ( t ), y 2 = y 2 ( t ), and Y = Y ( t ). First we find the functions y 1 and y 2 as solutions to the homogeneous equation y 3 y 4 y = 0 . We found above that a fundamental set of solutions is given by y 1 ( t ) = e t and y 2 ( t ) = e 4 t . Second we seek a function Y = Y ( t ) as a solution to the nonhomogeneous equation Y 3 Y 4 Y = 2 sin t. It su ces just to find one solution Y = Y ( t ). To this end, we make a guess that such a solution is in the form Y ( t ) = A cos β t + B sin β t for some constants A , B , and β . We have the derivatives Y = A cos β t + B sin β t Y = β B cos β t + β A sin β t Y = β 2 A cos β t + β 2 B sin β t This gives the expression Y 3 Y 4 Y = β 2 A 3 β B 4 A cos β t + β 2 B + 3 β A 4 B sin β t = 2 sin t. We now choose β = 1, so that 2 sin t = Y 3 Y 4 Y = 5 A 3 B cos t + 3 A 5
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