Interpreting the Constant c in a Savings Account Differential Equation

[David Donald]

Hello! so I'm working on a homework problem where I need to
solve the diff equation dy/dt = ky - W
where Y is the amount of money in a savings account
W is a constant
k is another constant (growing at a percentage)
solving the diff equation

I got y = W(1/k) + ((c)/(e^-kx))

now my question is how do i interpret the constant c? what does the constant c mean?

 

[blue_leaf77]

(e^-kx)

That should be $t$.

what does the constant c mean?

$c$ can be determined by the knowledge about the amount of the saving at a particular time.

 

[HallsofIvy]

"y" is "amount of money" in the account. It has units of, say, "dollars". "dy/dt" is the rate at which money goes into or out of the account. It has units of, say, "dollars per day". In order to have dy/dt= ky+ W, W must have units of "dollars per day" and k must have units of "per day" or "1/day".

Then, in y= W/k+ ce^(kt) (You have "c/e^(-kx)". Obviously, your "x" should be "t" and 1/e^(-kt) = e^(kt).)
Since k has units of "per day" and W "dollars per day", W/k has units of "dollars" as does y. Also t, the time, has units of "day" kt is "dimensionless" and e^(kt) also has no dimensions. That means that c must have dimensions of "dollars". Further, taking t= 0, we have y(0)= W/k+ c. c= y(0)- W/k. c is the amount of money in the account, initially, minus W/k. You don't say what "W" and "k" mean so I can't say what y(0)- W/k means.

 

[pasmith]

"y" is "amount of money" in the account. It has units of, say, "dollars". "dy/dt" is the rate at which money goes into or out of the account. It has units of, say, "dollars per day". In order to have dy/dt= ky+ W, W must have units of "dollars per day" and k must have units of "per day" or "1/day".

Then, in y= W/k+ ce^(kt) (You have "c/e^(-kx)". Obviously, your "x" should be "t" and 1/e^(-kt) = e^(kt).)
Since k has units of "per day" and W "dollars per day", W/k has units of "dollars" as does y. Also t, the time, has units of "day" kt is "dimensionless" and e^(kt) also has no dimensions. That means that c must have dimensions of "dollars". Further, taking t= 0, we have y(0)= W/k+ c. c= y(0)- W/k. c is the amount of money in the account, initially, minus W/k. You don't say what "W" and "k" mean so I can't say what y(0)- W/k means.

It is easy enough to interpret $k$ as a (compound) interest rate and $W$ as a constant rate at which funds are withdrawn from the account.

Arbitrary constants of integration (such as $c$) often don't have any good interpretation in terms of the application. However, their values are completely forced by requirements which come from the application. Here, for example, you might be interested in what the initial balance must be if the balance of the account is to be zero at some fixed time, or you might be interested in whether the balance will ever become zero given a specific initial balance.

 

[David Donald]

"y" is "amount of money" in the account. It has units of, say, "dollars". "dy/dt" is the rate at which money goes into or out of the account. It has units of, say, "dollars per day". In order to have dy/dt= ky+ W, W must have units of "dollars per day" and k must have units of "per day" or "1/day".

Then, in y= W/k+ ce^(kt) (You have "c/e^(-kx)". Obviously, your "x" should be "t" and 1/e^(-kt) = e^(kt).)
Since k has units of "per day" and W "dollars per day", W/k has units of "dollars" as does y. Also t, the time, has units of "day" kt is "dimensionless" and e^(kt) also has no dimensions. That means that c must have dimensions of "dollars". Further, taking t= 0, we have y(0)= W/k+ c. c= y(0)- W/k. c is the amount of money in the account, initially, minus W/k. You don't say what "W" and "k" mean so I can't say what y(0)- W/k means.

W is an amount that is getting withdrawed each year
and K is a percentage based growth

 

[haruspex]

W is an amount that is getting withdrawed each year
and K is a percentage based growth

As pasmith wrote, the constant is what allows you to fit the solution to the boundary conditions in the actual problem.
But check the signs in your solution. If you substitute it back into the differential equation does it look right?