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**tjsdndnjs** Verify that the given function is a solution of the DE and A and B are constants.

(a) y’’ + y = 0; y = A cos x + B sin x

How do I start this problem??? I have no clue where to start. Thank you.

We have

y''+y=0

And a statement that says that y is a solution where

y = Acos(x)+Bsin(x)

Which implies that

y = Acos(x)+Bsin(x) => y' = -Asin(x)+Bcos(x) => y'' = -Acos(x)-Bsin(x)

If we go back to y''+y'=0 we can check if the left side equals 0 when using the given derivatives of the function above. Hence

y''+y = -Acos(x)-Bsin(x)+Acos(x)+Bsin(x) = 0

Which solves the DE and hence Q.E.D