Science:Math Exam Resources/Courses/MATH105/April 2011/Question 05 (b)
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Question 05 (b) 

You open a savings account under the Escalator Plan with an initial deposit of . The advantage of this plan is that the interest rate is not fixed, but grows proportionally with time as long as the account is alive. In other words, the money in the account collects interest at the annual rate of at time compounded continuously (here is a constant). You also keep adding money to the account in the form of a continuous deposit of at time Now suppose that , , and Under these assumptions, solve the initial value problem you wrote down in part (a). 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

What type of differential equation do you obtain? 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. We now know that for all and as well . Thus, our differential equation specializes to with the initial condition . This is a separable differential equation that can be written as The first step to solve this differential equation is to integrate both sides of the equation with respect to : Informally, in the integrand of the integral on the lefthand side the terms cancel out each other. We arrive at The integral on the lefthand side is just for a constant . Since from the context is always nonnegative, we can ignore the absolute value brackets. In addition, we can combine the two constants and to a new constant that we denote by . We arrive at Applying the exponential function on both side of the equation results in We use the initial condition to determine the constant : Hence, the solution to the initial value problem is given by the function 