The differential equation for `y=Acos alphax +B sin alphax`, where A and B are arbitary constant is

Options

A. `(d^(2)y)/(dx^(2))-alpha .^(2)y=0`
B. `(d^(2)y)/(dx^(2))+alpha .^(2)y=0`
C. `(d^(2)y)/(dx^(2))+alphay=0`
D. `(d^(2)y)/(dx^(2))-alphay=0`

Answer

B

Explanation

Given, `y=A cos alpha+B sin alpha`
`Rightarrow (dy)/(dx)=-alpha A sin alpha x+alphaBcos x`
Again, differentiating both sides, w.r.t. x, we get
`Rightarrow(d^(2)y)/(dx^(2))=-Aalpha^(2)(Acos alphax+Bsin alphax)`
`Rightarrow(d^(2)y)/(dx^(2))=-alpha^(2)y`
`Rightarrow(d^(2)y)/(dx^(2))+alpha^(2)y=0`