91Ó°ÊÓ

To derive a differential equation for the loan principal P(t), we will first break down the loan repayment process into smaller parts. The homebuyer owes \(\ P_{0}\) dollars at an annual interest rate r, and agrees to repay the loan with equal monthly payments of M dollars per month over N years. The change in principal (loan amount) in a small time interval dt can be expressed as the difference between the interest incurred and the monthly payment made, i.e., \(\frac{rP(t)}{12}dt - M dt\). Using this relationship, we can write the differential equation for the loan principal P(t) at time t > 0: \(\frac{dP}{dt} = \frac{rP(t)}{12} - M\)

To solve the given differential equation, we rewrite it in the form of a first order linear differential equation: \(\frac{dP}{dt} - \frac{r}{12}P(t) = -M\) We can then solve this equation using an integrating factor, which for a linear differential equation of the form \(\frac{dy}{dt} + ky = f(t)\) is given by \(I(t) = e^{\int k dt}\). In this case, k = \(\frac{r}{12}\) and so, \(I(t) = e^{\int \frac{r}{12} dt} = e^{\frac{rt}{12}}\) Multiplying the differential equation by the integrating factor, we get: \(e^{\frac{rt}{12}}\frac{dP}{dt} - \frac{r}{12}e^{\frac{rt}{12}}P(t)=-Me^{\frac{rt}{12}}\) Now observe that the left-hand side is the derivative of the product \(P(t)e^{\frac{rt}{12}}\), so we can integrate both sides with respect to t: \(\int\frac{d}{dt} [P(t)e^{\frac{rt}{12}}] dt = \int -M e^{\frac{rt}{12}} dt\) \(P(t)e^{\frac{rt}{12}} = -\frac{12M}{r} e^{\frac{rt}{12}} + C\) On re-arranging, we get the general solution for P(t): \(P(t) = -\frac{12M}{r}e^{-\frac{rt}{12}} + Ce^{-\frac{rt}{12}}\) Applying the initial condition, \(P(0) = P_{0}\), we can find the value of the constant C: \(P_{0} = -\frac{12M}{r} + C\) \(C = P_{0} + \frac{12M}{r}\) Thus, the particular solution for P(t) is: \(P(t)=-\frac{12M}{r}e^{-\frac{rt}{12}}+ (P_{0}+\frac{12M}{r})e^{-\frac{rt}{12}}\)

To find the approximate value of the monthly payment M, we will first assume that the homebuyer repays the loan amount in exactly N years, i.e., \(P(t) = 0\) when \(t = 12N\). Applying this condition, we have: \(0=-\frac{12M}{r}e^{-\frac{r(12N)}{12}}+ (P_{0}+\frac{12M}{r})e^{-\frac{r(12N)}{12}}\) Solving this equation for M, we get: \(M \approx \frac{rP_{0}}{12}\left[1 - e^{-N r}\right]^{-1}\)

We now compare the value of M obtained using the approximate equation with the given exact value when (i) \(P_{0}=\$50,000\), \(r=7 \frac{1}{2}\%\), \(N=20\) and (ii) \(P_{0}=\$150,000\), \(r=9.0\%\), \(N=30\). (i) For \(P_{0}=\$50,000\), \(r=7.5\%\), \(N=20\): Approximate M value: \(M \approx \frac{0.075(50,000)}{12}\left[1 - e^{-20(0.075)}\right]^{-1} \approx \$449.71\) Exact M value: \(M = \frac{0.075(50,000)}{12}\left[1 - (1 + 0.075/12)^{-12 * 20}\right]^{-1} \approx \$454.89\) (ii) For \(P_{0}=\$150,000\), \(r=9\%\), \(N=30\): Approximate M value: \(M \approx \frac{0.09(150,000)}{12}\left[1 - e^{-30(0.09)}\right]^{-1} \approx \$1217.38\) Exact M value: \(M = \frac{0.09(150,000)}{12}\left[1 - (1 + 0.09/12)^{-12 * 30}\right]^{-1} \approx \$1247.00\) As we can see, the approximate values of M are fairly close to the exact values, indicating that the continuous repayment model provides a reasonable approximation to the actual loan repayment process.