Susan borrows $A$ dollars at an annual interest rate of $5$ percent. Anticipating steady salary increases, she expects to make payments at a monthly rate of $900(\dfrac{1+t}{200})$ dollars per month, where $t$ is the number of months since the loan was made. Assume that interest is compounded continuously and the payments are also made continuously. Let $y$ be the amount of money that Susan still owes $t$ months after the loan is made. Write a differential equation that models $y$.

The following is the equation that I set up, where I divided the annual rate of $5$ percent by 12 to convert it into monthly rate. $$\dfrac{dy}{dt}= A +(1+\dfrac{0.05}{12})y-900(\dfrac{1+t}{200})$$

But this is not the right answer, can anyone point out where I did wrong?

  • $\begingroup$ Why you didn't multiply $$900\frac{1+t}{200}$$ with $y$? Because she returns that amount of money per month. $\endgroup$ – Stefan4024 Sep 27 '13 at 2:15
  • $\begingroup$ @Stefan4024, but $y$ is the amount she owes, and $900\dfrac{1+t}{200}$ is the amount she returns. What's the point of multiply these two things? $\endgroup$ – user59036 Sep 27 '13 at 2:22
  • $\begingroup$ I misunderstand you there, I thought that $y$ is the time. $\endgroup$ – Stefan4024 Sep 27 '13 at 2:30

You should not have the $A$ alone on the right. $y(0)=A$ is the initial condition, but the interest is applied to $y$. The differential equation comes from $\frac {dy}{dt}=$interest-payments, so $\frac {dy}{dt}=\frac {5\%}{12}y-900\left(\frac{1+t}{200}\right)=\frac y{240}-\frac 92(1+t)$


I think that you problem is that the generating formula is wrong.

Let $y(t)$ represents the money she owes at given moment $t$

We know that she borrowed $A$ money.

We know that the annual interest is $5\%$, so the montly interest will be $\frac{5\%}{12}$

So after $t$ months the total amount of money will be:

$$(A + \frac{5\% A}{12})t = (A + \frac{\frac{5A}{100}}{12})t = (1 + \frac{1}{240})At$$

But she return a certain amount of money and that's of rate: $900\frac{1+t}{200}$

So the formula will finally reach its final look and it'll be:

$$y(t) = \left(1 + \frac{1}{240}\right)Ay - 9y \frac{1+t}{2}$$

Now you can write it as differential equation.

  • $\begingroup$ The interest compounds, so if there were no payments it would be $A(1+\frac 1{240})^t$. Each payment made should also have interest applied to it. Your last equation the units don't match-the term in Ay is (money)^2 while the others are (money). $\endgroup$ – Ross Millikan Sep 27 '13 at 2:56

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