lecture_15 (dragged) 1

lecture_15 (dragged) 1 - 2 MA 36600 LECTURE NOTES FRIDAY...

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

View Full Document Right Arrow Icon
2 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20 We mention in passing that we have the identity W 0 =det ° y 1 ( t 0 ) y 2 ( t 0 ) y ° 1 ( t 0 ) y ° 2 ( t 0 ) ± ° 10 01 ± =1 . Linear Independence Abel’s Theorem. Consider the diferential equation a ( t ) y °° + b ( t ) y ° + c ( t ) y =0 . Say that y 1 = y 1 ( t ) and y 2 = y 2 ( t ) are solutions. (1) The Wronskian oF y 1 and y 2 is the Function W ( t )= C exp ° ² t b ( τ ) a ( τ ) ± For some constant C . (2) W ( t ) ° = 0 For all t iF and only iF W ( t 0 ) ° = 0 For some t 0 . We give a prooF. Denote the Function W = y 1 y ° 2 y ° 1 y 2 . This has the derivative W ° = ³ y ° 1 y ° 2 + y 1 y °° 2 ´ ³ y °° 1 y 2 + y ° 1 y ° 2 ´ = y 1 y °° 2 y °° 1 y 2 . Hence we have the expression a ( t ) W ° + b ( t ) W = a ( t ) µ y 1 y °° 2 y °° 1 y 2 + b ( t ) µ y 1 y ° 2 y ° 1 y 2 = a ( t ) µ y 1 y °° 2 y °° 1 y 2 + b ( t ) µ y 1 y ° 2 y ° 1 y 2 + c ( t ) µ y 1 y 2 y 1 y 2 = y 1 µ a ( t ) y °° 2 + b ( t ) y ° 2 + c ( t ) y 2 µ a ( t ) y °° 1 + b ( t ) y ° 1 + c ( t ) y 1 y 2 . Hence W = W ( t ) satis±es a ±rst order diferential equation. Since this equation is separable, we may solve by dividing both sides by W ( t ): a ( t ) dW dt = b ( t ) W 1 W dW dt = b ( t ) a ( t ) d dt ln | W | = b ( t ) a ( t ) = ln | W | = ² t b ( τ ) a ( τ ) + (constant) . Upon exponentiating, we can express W = W ( t ) in the Form
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online