Determine whether the given function is a solution to the given differential equation. x =2cost - 3sint, x'' +x =0

Given :

A function , x = 2cos t -3sin t               .....equation 1.

A differential equation , x'' + x = 0      .....equation 2.

To Find :

Whether the given function is a solution to the given differential equation.

Solution :

First derivative of x :

Now , second derivative :

( Note : derivative of sin t is cos t and cos t is -sin t )

Putting value of x'' and x in equation 2 , we get :

=(-2cos t + 3sin t ) + ( 2cos t -3sin t )

= 0

So , x'' and x satisfy equation 2.

Therefore , x function is a solution of given differential equation .

Hence , this is the required solution .

i think it might be a but you might want a second opinion