You want to save $20,000 to buy a new car four years from now. You decide to ope

Question

You want to save $20,000 to buy a new car four years from now. You

decide to open an account for this purpose, intially with no money in

it, and you plan to deposit a fixed amount of money each month.

(a)

Write a differential equation that describes the rate of growth dp/dt

of the principal in the account assuming that interest and deposits are

added continuously.

(b)

Suppose the account earns interest at the rate of 6% per year.

How much money do you need to deposit each month in order to save

$20,000 in four years? (Assume a continous, ODE model, of course.)

Details please

Explanation / Answer

Suppose you add x $ at the end of each month

then at the end of 12 month, you will be adding 12x $ to the previous sum and its interest

let the current amount of money be denoted as X

Hence the rate of change at the end of each month = interest +12 x

say interest = y % per year

then dX/dt = 12x + y* X/100 (This is your differential equation)

=> 100 dX/dt = 100*12x + y * X

=> 100 dX / (100*12x + y*X) = dt

Let 100 * 12x + y*X = z

then dz = y dX

Substituting this we get

(100/y) * dz/z = dt

=> (100/y) * ln z = t + C

=> 100 ln z = yt + C [ constant * y = another constant]

=> ln z = yt/100 + C

or z = K . e^(yt/100) [ where K is another constant]

replace back z now

1200 x + yX = K e^(yt/100)

or X = (K e^(yt/100) - 1200x) / y

Now we need to solve for K

at the begining of time t=0, X = 0

so put that in our quation, we get

0 = K - 1200x or K = 1200x

So X = (1200x e^(yt/100) - 1200x) / y

X = 1200x * ( e^(yt/100) - 1) / y

This is the model that will give you the current sum at the end of a year

(b) X = 20000 and y = 6

then 20000 = 1200x * (e^(0.06 * 4) - 1) / 6

=> 100 = x * (e^(0.06 * 4) - 1)

=> 100 = x * 0.2712

=> x = 100 / 0.2712 = 368.73 $ per month

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