Question

Determine whether the given function is a solution to the given differential equation.

Answer 1

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 :

$x'=\dfrac{d(2cos t - 3sin t)}{dt}\\\\x'=\dfrac{d(2cost)}{dt}-\dfrac{(3sint)}{dt}\\\\x'=-2sint-3cost$

Now , second derivative :

$x''=\dfrac{d(-2sint-3cost)}{dt}\\\\x''=-\dfrac{d(2sint)}{dt}-\dfrac{d(3cost)}{dt}\\\\x''=-2cost+3sint$

( 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 .