Respuesta :
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 :
[tex]x'=\dfrac{d(2cos t - 3sin t)}{dt}\\\\x'=\dfrac{d(2cost)}{dt}-\dfrac{(3sint)}{dt}\\\\x'=-2sint-3cost[/tex]
Now , second derivative :
[tex]x''=\dfrac{d(-2sint-3cost)}{dt}\\\\x''=-\dfrac{d(2sint)}{dt}-\dfrac{d(3cost)}{dt}\\\\x''=-2cost+3sint[/tex]
( 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 .
