jayaprakashpadyala

A certain aqueous solution of sucrose at 25oC has a vapor pressure of 23.34 torr (a unit for pressure). The vapor pressure of water at 25oC is 23.76 torr. Find the activity of the solvent water in this sucrose solution. 

In this problem, we use the method of separation of variables to find the family of solutions
to the very important differential equation dP/dt = kP , for constant k. We have seen that
P (t) = Cekt is a family of solutions where C is any real number. We got this via ‘guess and
check.’ Now, we will see how the method of separation of variables and careful reasoning
leads to this. (Note that it’s not the only way to get the result. Guess and check is a valid
method.)

(a) Use the method of separation of variables to find a (family of) implicit solution(s) to
this differential equation. (Don’t forget the absolute value in the antiderivative of 1/P
and use D for your constant of integration.)

(b) Now, begin to solve for P by exponentiating both sides.

(c) Use exponent laws to write the formula you have for |P (t)| as a product.

(d) What type of number is eD, for any D? (Any real number? Positive? Negative?
Nonpositive? Greater than 1? etc.)

(e) Rename eD as A but add that A ∈ (?, ?), with the question marks replaced by the
appropriate values or symbols. (Use (d) above.)

(f) If |P (t)|= f (t) for some function f (t) > 0 for all t (and if P (t) is continuous, which it
is), then either P (t) = f (t) for all t or P (t) = −f (t) for all t. Often, we write this as
P (t) = ±f (t). Use this to remove the absolute values from your result for (e).

(g) Rename ±A as B but add B ∈ (?, ?) or B ∈ (?, ?) with the question marks replaced
by the appropriate values or symbols.

(h) Show that P (t) = 0 for all t is a solution to this differential equation.

(i) Finally, combine your result from (g) with P (t) = 0 to write the family of solutions
as P (t) = Cekt with C ∈ (?, ?) with the question marks replaced by the appropriate
values or symbols..

Suppose a,b,c,s,t are all integers, a 6=0,a∣b and a∣c . Then a∣(sb+tc) . Use it in a proof of the following: Prove that if a is a non- 0 integer, k∈Z,a∣(4k+3) , and a∣(2k+1) , then a=1 or a=−1 . Suggestion: use proof above and the fact that only divisors of 1 are 1 and -1 .