lecture_18 (dragged) - ) or some unction v = v ( t ) to be...

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MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Repeated Roots Review. Say that we wish to fnd the general solution to the diFerential equation ay °° + by ° + cy =0 where b 2 4 ac = 0. ±irst note that the roots o² 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 } ²orms a ²undamental set o² solutions, where y 1 ( t )= e rt and y 2 ( t )= te rt . Example. Consider the homogeneous diFerential equation y °° +4 y ° +4 y =0 . We fnd the general solution using d’Alembert’s Method. ±irst we fnd one solution in the ²orm 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 ²orm y ( t )= v ( t ) y 1 ( t
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Unformatted text preview: ) or some unction v = v ( t ) to be ound. We have the derivatives y = v e 2 t y = ( v 2 v ) e 2 t y = ( v 4 v + 4 v ) e 2 t This gives the expression 0 = y + 4 y + 4 y = ( v 4 v + 4 v ) e 2 t + 4 ( v 2 v ) e 2 t + 4 v e 2 t = v e 2 t so that v = 0. We know that the general solution to this diFerential equation is v ( t ) = c 1 + c 2 t or constants c 1 and c 2 . Hence the general solution to the diFerential equation y + 4 y + 4 y = 0 is the unction y ( t ) = c 1 e 2 t + c 2 t e 2 t . 1...
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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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