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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 .
for this case we have to define trigonometric relations of rectangular triangles, that the tangent of an angle is given by the leg opposite the angle on the leg adjacent to the angle. So:
Answer: