{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_18 (dragged)

# lecture_18 (dragged) - ²or some ²unction v = v t to be...

This preview shows page 1. Sign up to view the full content.

MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Repeated Roots Review. Say that we wish to find the general solution to the di ff erential equation a y + b y + c y = 0 where b 2 4 a c = 0. First note that the roots of the characteristic polynomial a r 2 + b r + c = 0 are r 1 = r 2 = b 2 a ± b 2 4 a c 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 ) = t e rt . Example. Consider the homogeneous di ff erential equation y + 4 y + 4 y = 0 . We find the general solution using d’Alembert’s 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) ²or some ²unction v = v ( t ) to be ²ound. 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 ²or constants c 1 and c 2 . Hence the general solution to the diFerential equation y ° + 4 y ° + 4 y = 0 is the ²unction y ( t ) = c 1 e − 2 t + c 2 t e − 2 t . 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online