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If y1(t) and y2(t) are two solutions to
y′′ + p (t ) y′ + q (t ) y = 0
So the Wronskian of the two solutions is,
W(y1,y2)(t) =
= for some t0.
Since we don't know the Wronskian and we don't know t0 this won't do us many good apparently. Though, we can rewrite as
W(y1,y2)(t) = ce-∫p(t) dt ...................................(3)
Here the original Wronskian sitting opposite the exponential is absorbed in the c and the evaluation of the integral at t0 will place a constant in the exponential such can also be brought out and absorbed in the constant c. Whether you don't recall how to do this return and take see the linear, first order differential equation section that we did something the same there.
Along with this rewrite we can calculate the Wronskian up to a multiplicative constant, that isn't too bad. See as well that we don't in fact need the two solutions to do that. All we require is the coefficient of the first derivative from the differential equation and provided the coefficient of the second derivative is one also.
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