aaa Exercise: Consider the second order linear differential equation: y\"+P(x)y\

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Exercise: Consider the second order linear differential equation: y"+P(x)y' + Q(x)y = 0. Show by direct substitution that if y, (x) is a solution of this differential equation then y2(x) = y1(x) is also a solution. This exercise illustrates a method of finding a second solution from a solution that is already known. Hint: Use the product, quotient and chain rules of differentiation to calculate y' and y" , the 1st and 2nd derivatives, implicitly, with respect to x, and then plug the results directly into the differential equation: y" +P(x)y' + Q(x)y = 0.

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