lecture_18

lecture_18 - MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27...

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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Repeated Roots Review. Say that we wish to find the general solution to the differential equation ay 00 + by + cy = 0 where b 2- 4 ac = 0. First note that the roots of the characteristic polynomial ar 2 + br + c = 0 are r 1 = r 2 =- b 2 a b 2- 4 ac 2 a =- b 2 a . Then { y 1 ,y 2 } forms a fundamental set of solutions, where y 1 ( t ) = e rt and y 2 ( t ) = te rt . Example. Consider the homogeneous differential equation y 00 + 4 y + 4 y = 0 . We find the general solution using dAlemberts Method. First we find one solution in the form y 1 ( t ) = e rt . Indeed, the characteristic equation is r 2 + 4 r + 4 = 0 = ( r + 2) 2 = 0 . Hence r 1 = r 2 =- 2 is the only root, so one solution is y 1 ( t ) = e- 2 t . Assume that the general solution is in the form y ( t ) = v ( t ) y 1 ( t ) for some function v = v ( t ) to be found. We have the derivatives y = v e- 2 t y = ( v- 2 v ) e- 2 t y 00 = ( v 00- 4 v + 4 v ) e- 2 t This gives the expression 0 = y 00 + 4 y + 4 y = ( v 00- 4 v + 4 v ) e- 2 t + 4( v- 2 v ) e- 2 t + 4 v e- 2 t = v 00 e- 2 t so that v 00 = 0. We know that the general solution to this differential equation is v ( t ) = c 1 + c 2 t for constants c 1 and c 2 . Hence the general solution to the differential equation y + 4 y + 4 y = 0 is the function y ( t ) = c 1 e- 2 t + c 2 te- 2 t . 1 2 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Summary. We summarize the general solution to the differential equation ay 00 + by + cy = 0 . Say that the characteristic equation ar 2 + br + c = 0 has roots r 1 and r 2 . A fundamental set of solutions for this constant coefficient differential equation is { y 1 ,y 2 } in terms of the functions y 1 ( t ) = e r 1 t if b 2- 4 ac > 0, e r 1 t if b 2- 4 ac = 0, e t cos t if b 2- 4 ac < 0....
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lecture_18 - MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27...

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