91Ó°ÊÓ

In our problem, a differential equation is used to describe how the balance in a savings account changes over time when continuously compounded interest and annual deposits are involved. A differential equation is a mathematical equation that relates a function with its derivatives. In this specific scenario, the equation is \( \frac{dn}{dt} = A + 0.05n \), where \(n\) is the amount in the account, \(A\) is the annual addition of 2000 dollars, and 0.05 represents the 5% interest rate being compounded continuously. This equation tells us how the rate of change of \(n\) with respect to time \(t\) depends on the current amount \(n\) and the additional constant income \(A\). By solving the differential equation, we can determine \(n\), the total amount in the account after a given number of years.

Compound interest is a method of accruing interest where interest earned is reinvested, allowing the investment to grow at an increasing rate. When compounded continuously, the interest is added to the account balance at every possible division of time, theoretically at every instant. This means it is not accumulated at regular intervals, such as annually or monthly, but constantly. In the problem's differential equation, the term \(0.05n\) represents this continuous interest.

• If you picture compound interest as a train accelerating over time, then continuously compounded interest is like a train never needing to stop accelerating.
• This involves using the exponential function to model the growth of investments or savings accounts due to continuous compounding.

Applying this concept lets us evaluate how quickly an account balance can grow when interest is constantly reinvested back into the account.

Integration is the reverse process of differentiation and it is used to solve differential equations. In our exercise, we have used an integration technique known as the integrating factor method. This method simplifies solving linear differential equations. An integrating factor is a function, in our case \(e^{0.05t}\), which, when multiplied by the differential equation, allows it to be rewritten so the left side forms a derivative of a product as in:

• \(e^{0.05t} \frac{dn}{dt} - 0.05e^{0.05t}n = 2000e^{0.05t}\).

This makes it easy to integrate both sides with respect to time \(t\). The left side becomes the derivative of \(e^{0.05t}n\), and on integrating this and the right side through substitution or direct integration, we can express \(n\) explicitly as a function of \(t\). This technique is particularly important when dealing with equations involving exponential growth or decay.

Calculating interest, especially with continuous compounding and additional annual contributions, can sound complex. However, by breaking it into parts using formulas derived from the integration of our differential equation, it becomes clear.

• Start with your base equation \(n = 40000 + Ce^{-0.05t}\), derived from solving the differential equation.
• Next, apply any initial conditions, such as \(n(0) = 0\), to determine constants like \(C\).
• Finally, substitute the specific time \(t\), like 5 years in this problem, back into the equation to find \(n(t)\), the value of the account at that time.

Using these steps with known numerical values like \(e^{-0.25} \approx 0.7788\), we calculate the account balance at 5 years as approximately 8848 dollars. This approach clearly lays out how much money accumulates over time with continuous compounding and regular additions.