Which of these numbers is four less than a square number?
A 3
B 20
C 37
D 45

45

Step-by-step explanation:

basically what we are trying to find is x + 4= square

for you to understand this better, take eg 5.

5  +   4 =       9

         |            \

four less than the square number( 9 is the square of 3)

we took the square of 3 and subtracted 4 from it making the result 4 less

12 + 4 = 16

         |            \

four less than the square number( 16 is the square of 4)

we took the square of 4 and subtracted 4 from it making the result 4 less

understanding that,

a)3

3 + 4 = 7

there is no square root of 7   X

b)20

20 + 4 = 24

there is no square root of 24   X

c)37

37 + 4 = 41

there is no square root of 41   X

d) 45

45 + 4 = 49

the square root of 49 = 7

answer=49

You're looking for a solution of the form

Differentiating twice yields

Substitute these series into the DE:

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

which indicates that the coefficients in the series solution are governed by the recurrence,

Use the recurrence to get the first few coefficients:

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

Just to confirm this solution is valid: we have

y = 10x - 7eˣ   ==>   y (0) = 0 - 7 = -7

y' = 10 - 7eˣ   ==>   y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.