Interest Compound Equation given by; y\'(t,y)=ry(t)-12*p, y(0)=y0 IO Mortgage Eq

Question

Interest Compound Equation given by;

y'(t,y)=ry(t)-12*p, y(0)=y0

IO Mortgage Equation given by

y'(t,y)=r(t)y(t)-12*p(t)

1. Are there any equilibrium solutions to the equations above. What do the equilibria represent?
2. Solve the Interest Compound equation using separation of variables, with y(0)=300000, r-.05

3. Use the solution to the differential equation for compound interest to correct p for a 10 year fixed rate mortgage, with rate 5% and initial debt 300000. Do the same for a 30 year fixed rate mortgage with rate 7%

4.Determine total paid my summing each monthly payment. How much interest is paid in a 30 year fixed rate mortgage? And the 10 year?
5. How much money could the borrow save in each case if he/she paid $50000 down on the house (Mortgage is now $250000)
6. What are the advantages and disadvantages of taking out a 30 year fixed rate mortgage as opposed to 10year?

Explanation / Answer

1. To find equlibrium solution set y' = 0, meaning that ry-12p = 0 so y = 12p/r. This means that if you start with 12p/r dollars in the account and draw out $p per month, the amount in the account will stay constant forever.

2. y' - ry = -12p so [e^(-rt) y]' =-12pe^(-rt) so e^(-rt) y = (12p/r)e^(-rt) + C so y = (12p/r) + Ce^(rt)

With r=0.05 we have y = 240p+Ce^(0.05t) and when t = 0, 300000 = 240p + C so C = 300000 - 240p

solution is: y = 240p+[300000 - 240p]e^(0.05t)

3. y(10) = 0 so 0 = 240p+[300000 - 240p]e^(0.05*10) means:

240p[e^(0.5)-1] = 300000e^(0.05*10)

so p = 1250e^(0.05*10)/[e^(0.5)-1] = $3176.87

With r = 0.07 and y(30)=0:

y = (12p/r) + Ce^(rt) = (12p/0.07)+Ce^(0.07t)

300000 = (12p/0.07)+Ce^(0) = (12p/0.07)+C so C = 300000-12p/0.07

y = (12p/0.07) + [300000-(12p/0.07)]e^(0.07t)

(12p/0.07)[e^(2.1)-1] = 300000e^(2.1)

p = 0.07*300000e^(2.1)/[12(e^(2.1)-1)] = $1994.20

4. 30-year: $1994.20*30*12 = $717913.07

10-year: $3176.87*10*12 = $381224.11

5. 30-year: p = 0.07*250000e^(2.1)/[12(e^(2.1)-1)] = $1661.84

savings: [$1994.20-$1661.84]*30*12 = $119649.60

10-year: p = [250000/240]e^(0.05*10)/[e^(0.5)-1] = $2647.39

savings: [$3176.87-$2647.39]*10*12 = $63537.60

6. advantages: lower monthly payments now and throughout loan

disadvantages: pay way more in interest over course of loan

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