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MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 23 3 is nonzero then y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) is the solution of the initial value problem in terms of the constants c 1 = y 0 y 2 ( t 0 ) y 0 y 2 ( t 0 ) W 0 and c 2 = y 0 y 1 ( t 0 ) y 0 y 1 ( t 0 ) W 0 . But we saw above that W y 1 , y 2 ( t 0 ) = W 0 = 0 if and only if y 1 and y 2 are linearly independent. In particular, to check that two solutions y 1 = y 1 ( t ) and y 2 = y 2 ( t ) form the general solution, it su ces to check that their Wronskian W y 1 , y 2 ( t ) = y 1 ( t ) y 2 ( t ) y 1 ( t ) y 2 ( t ) is nonzero for some t = t 0 . In this case, we say that { y 1 , y 2 } forms a Fundamental Set of Solutions to the di ff erential equation. We have seen that if y 1 = y 1 ( t ) and y 2 = y 2 ( t ) are linearly independent solutions to some di ff erential equation in the form a ( t ) y + b ( t ) y + c ( t ) y = 0 then we can find the general solution y ( t ) = c 1 y 1 ( t )+ c 2 y 2 ( t ). We mention in passing that given two linearly independent functions f = f ( t ) and g = g ( t ), we can always find a homogeneous second order di ff erential equation y + p ( t ) y + q ( t ) y = 0 for which y ( t ) = c 1 f ( t ) +
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