The equations (2) and (3) you referred to are unavailable, but it is clear that you are trying to show that two set of solutions y1 and y2, to a (second-order) differential equation are solutions, and form a fundamental set. This will be explained.
Answer:
SOLUTION OF A DIFFERENTIAL EQUATION.
Two functions y1 and y2 are set to be solutions to a differential equation if they both satisfy the said differential equation.
Suppose we have a differential equation
y'' + py' + qy = r
If y1 satisfies this differential equation, then
y1'' + py1' + qy1 = r
FUNDAMENTAL SET OF DIFFERENTIAL EQUATION.
Two functions y1 and y2 are said to form a fundamental set of solutions to a second-order differential equation if they are linearly independent. The functions are linearly independent if their Wronskian is different from zero.
If W(y1, y2) ≠ 0
Then solutions y1 and y2 form a fundamental set of the given differential equation.
Let us consider the given expression:
in the scientific notation.
= 
[Using the law of exponents, which states
]
= 
On dividing, we get
= 
Hence, it is the required scientific notation of the given expression.
Answer:
x = -4, 10
Step-by-step explanation:
We take the square root and use the fact of "plus-minus" and find two values of x. The process of solving this equation is shown below:
Hence the two values of x are -4 and 10
Note: both (-7)^2 and (7)^2 are 49 . So we took "plus-minus" values of 
Answer:
C
Step-by-step explanation:
This is a combination question.
In the first instance, we select 5 from 20 and in the second case , we select 4 from 20.
The total number of ways to solve the first instance is 20C5 = 15504 ways
The total number of ways to solve the second instance is 20C4 = 4,845
The ratio of the first to the second scenario is 15,504/4,845 = 3.2 = 16 to 5
