1.8  Linear Equations

September 18, 1998

A Finance Problem

A student has saved $30,000 for her college tuition.  When she starts college, she invests the money in a savings account earning 6% interest per yer, compounded continuously.  Suppose her college tuition is $10,000 per year the first year, increases $500 per year per year after the first, and she arranges with her college that the money will be deducted from her savings account in small payments.  In other words, we assume that she is paying continuously.  How long will she be able to stay in school until she runs out of money?

Setup

t = time (in years) after she starts college

A = amount (in $1000) in her savings account

dA/dt = Float(6, -2)*A-(10+Float(5, -1)*t) , A(0) = 30 .

Solving the DE

> dsolve({diff(A(t),t)=.06*A(t)-(10+.5*t),A(0)=30},A(t));

A(t) = 305.5555556+8.333333333*t-275.5555556*exp(0.6000000000e-1*t)

Solving the problem

The question is, when does she run out of money, i.e., when does her savings account reach zero?

> solve(305.5555556+8.333333333*t-275.5555556*exp(.6000000000e-1*t)=0);

-31.74375314, 3.057040835

If this answer seems a little cryptic, we can plot the function A(t) .

> plot(305.5555556+8.333333333*t-275.5555556*exp(.6000000000e-1*t),
t=-32..4);

[Plot]

This shows the two zeroes.  We are, of course, only interested in positive times t .

> plot(305.5555556+8.333333333*t-275.5555556*exp(.6000000000e-1*t),
t=0..4);

[Plot]

The final answer:  She runs out of money after Float(3057040835, -9) years.