We obtain a loan of initial value L at an interest rate of r

We obtain a loan of initial value L at an interest rate of r per year and make payments of p per month. The amount A of the loan (approximately) satisfies the equation dA/dt = rA/12 p , A(0) = P. Here time t is measured in months. a) Solve for A(t). b) Let P = 30,000 and r = 2%. What value of p is required to pay of the loan in 5 years?

Solution

dA/dt = rA/12 p

a) Solve for A(t) , solve the differential equation:

dA/( rA/12 -p) = dt

Integrating both sides we get:

12/r ln( rA/12 -p) = t + C1 where C1 is a constant

( rA/12 +p) = e^(rt/12+C2)

rA = {e^(rt/12+C2) -p }

A =  {e^(rt/12+C2) -p }/r

A = -p/r + C3 e^(rt/12)

A = C3 e^(rt/12) - p/r

t=0 A(0) =P Solve for constant C3

p = C3 -p/r

C3 = p+p/r = (pr+p)/r

A = =(pr+p)/r e^(rt/12) - p/r

P = 30, 000 r = 2% t = 5yrs . Solve for A(t)

A(t) = {( 30000*2 +30000)/2}e^(2*5/12) - 30000/2 }

= $ 88 543.91