Question

A person borrows $10,000 and repays the loan at the rate of $2,400 per year. The lender charges interest of 10% per year. Assuming the payments are made continuously and interest is compounded continuously (a pretty good approximation to reality for long-term loans), the amount M(t) of money (in dollars) owed t years after the loan is made satisfies the differential equationdMdt = 110 M − 2400and the initial conditionM(0) = 10000.(a) Solve this initial-value problem for M(t).M(t) = (b) How long does it take to pay off the loan? That is, at what time t is M(t) = 0? Give your answer (in years) in decimal form with at least 3 decimal digits. years

Answer 1

Answer:

a) $M(t)=24000-14000e^{t/10}$

b) It takes 5.390 years to pay off the loan.

Step-by-step explanation:

(a) The differential equation for the variation in the amount of the loan M is

$\frac{dM}{dt}=0.1M-2400$

We can express this equation as:

$\frac{dM}{dt}=0.1M-2400\\\\10\int \frac{dM}{M-24000} =\int dt\\\\10ln(M-24000)=t\\\\M-24000=Ce^{t/10}\\\\M=Ce^{t/10}+24000\\\\M(0)=Ce^{0/10}+24000=10000\\\\C+24000=10000\\\\C=-14000\\\\M(t)=24000-14000e^{t/10}$

(b) We can calculate this as M(t)=0

$M(t)=24000-14000e^{t/10}=0\\\\e^{t/10}=24000/14000=1.714\\\\t=10ln(1.714)=10* 0.5390=5.390$

It takes 5.390 years to pay off the loan.

Answer 2

A) The initial Value problem for M(t) is; M(t) = 24000 - 14000e^(t/10)

B) The time that it will take for the person to pay off the loan is; 5.388 years

How to solve differential equations?

A) We are given the differential equation for the amount M(t) of money (in dollars) owed t years after the loan is made as;

dM/dt = 0.1M - 2400

Let us rearrange to get;

dM/(0.1M - 2400) = dt

Integrate both sides;

∫dM/(0.1M - 2400) = ∫dt

10In (M - 24000) = t

M - 24000 = Ae^(t/10)

M = 24000 + Ae^(t/10)

We are given initial condition of M(0) = 10000. Thus;

M(0) = 24000 + Ae^(0/10) = 10000

24000 + A = 10000

A = 10000 - 24000

A = -14000

Thus, the initial value problem is;

M(t) = 24000 - 14000e^(t/10)

B) We want to find the time at which M(t) = 0. Thus;

0 = 24000 - 14000e^(t/10)

24000 = 14000e^(t/10)

e^(t/10) = 24000/14000

e^(t/10) = 1.714

t/10 = In 1.714

t = 10 * 0.5388

t = 5.388 years

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