Originally Posted by
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