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

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MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20 Linear Homogeneous Equations Example. Consider the di ff erential equation y + 5 y + 6 y = 0 . We discuss the Wronskian associated with this equation. We have the characteristic equation r 2 + 5 r + 6 = 0 . Since the polynomial factors as r 2 = 5 r +6 = ( r + 2) ( r + 3), the characteristic equation has the two distinct real roots r 1 = 2 and r 2 = 3. Hence two solutions to the di ff erential equation are y 1 ( t ) = e 2 t and y 2 ( t ) = e 3 t . The Wronskian of y 1 and y 2 is the function W ( t ) = y 1 y 2 y 1 y 2 = e 2 t 3 e 3 t 2 e 2 t e 3 t = 3 e 5 t + 2 e 5 t = e 5 t . Example. Consider the more general constant coe cient equation a y + b y + c y = 0 . We have the characteristic equation a r 2 + b r + c = 0 . Say that this has roots r 1 and r 2 , and consider the solutions y 1 ( t ) = e r 1 t and y 2 ( t ) = e r 2 t . The Wronskian of y 1 and y 2 is the function W ( t ) = y 1 y 2 y 1 y 2 = e r 1 t r 2 e r 2 t r 1 e r 1 t e r 2 t = ( r 2 r 1 ) e ( r 1 + r 2 ) t .
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