Jorge got a hit 8/25 of the times that he
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If you raise a fraction to an exponent, you apply that exponent to the numerator and the denominator. This would give us .
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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 .