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# lecture_15 - MA 36600 LECTURE NOTES FRIDAY FEBRUARY 20...

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MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20 Linear Homogeneous Equations Example. Consider the differential equation y 00 + 5 y 0 + 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 differential 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 0 2 - y 0 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 coefficient equation a y 00 + b y 0 + 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 0 2 - y 0 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 . In particular, we have the value W 0 = W ( t 0 ) = ( r 2 - r 1 ) e ( r 1 + r 2 ) t 0 . Hence W 0 6 = 0 precisely when r 1 6 = r 2 . In fact, if r 1 6 = r 2 then W ( t ) 6 = 0 for any t . Fundamental Set of Solutions. We spend the remainder of the lecture discussing when solutions y 1 and y 2 form the general solution y = c 1 y 1 + c 2 y 2 . Consider the differential equation a ( t ) y 00 + b ( t ) y 0 + c ( t ) y = 0 . Say that we can find two solutions y 1 = y 1 ( t ) and y 2 = y 2 ( t ) satisfying the initial conditions y 1 ( t 0 ) = 1 y 0 1 ( t 0 ) = 0 ) and ( y 2 ( t 0 ) = 0 y 0 2 ( t 0 ) = 1 The set { y 1 , y 2 } is called a fundamental set of solutions . We will show that the general solution to the differential equation is y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) for constants c 1 and c 2 . To see why this is true, consider the initial conditions y ( t 0 ) = y 0 and y 0 ( t 0 ) = y 0 0 . We must construct constants c 1 and c 2 which solve this initial value problem. By the Principle of Super- position, we know that y = c 1 y 1 + c 2 y 2 is a solution to the differential equation. We have the system of equations y 0 = c 1 y 1 ( t 0 ) + c 2 y 2 ( t 0 ) = c 1 y 0 0 = c 1 y 0 1 ( t 0 ) + c 2 y 0 2 ( t 0 ) = c 2 Hence y ( t ) = y 0 · y 1 ( t ) + y 0 0 · y 2 ( t ) is the solution to the initial value problem.

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