The solution depends on the value of

. To make things simple, assume

. The homogeneous part of the equation is

and has characteristic equation

which admits the characteristic solution

.
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be

. Then

So you have


This means


and so the general solution would be
24 makes the equation true
Answer: 
Step-by-step explanation:
We know that probability for any event = 
Given : Charlotte has 6 cherry candies, 3 grape candies, and 3 lime candies.
I..e Total pieces of candies she has = 6+3+3= 12
Now , If Charlotte randomly pulls one piece of candy out of the bag, what is the probability that it will be cherry is given by :-

Hence, the probability that it will be cherry is
.
Answer:
x > − 7
Step-by-step explanation:
hope this helps
Answer:
Step-by-step explanation:
L.C.M. of 2 and 3=6
Raise to the power 6