First Order Differential Equation

In summary, the college graduate expects to make monthly payments of $800 at an interest rate of 9%. The loan will be fully paid in 135.36 months.
  • #1
DivGradCurl
372
0
"A recent college graduate borrows $100,000 at an interest rate of 9% to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800(1 + t/120), where t is the number of months since the loan was made. Assuming that this payment schedule can be maintained, when will the loan be fully paid?"

Please, help me find where I made a mistake. Here's what I've got:

[tex]S_0 = \$ 100,000 [/tex]

[tex]r = 0.09[/tex]

[tex]k(t) = \$ 800 \left( 1 + \frac{t}{120} \right) / \mbox{month}[/tex]

[tex]\frac{dS}{dt}=rS-k(t), \qquad S(0)=S_0[/tex]

[tex]\frac{dS}{dt}-rS=-k(t)[/tex]

[tex]\mu = \exp \left( -r \int dt \right) = e ^{-rt}[/tex]

[tex]S(t)=e^{rt}\int -800 \left( 1 + \frac{t}{120} \right) e ^{-rt} \: dt[/tex]

[tex]S(t)=e^{rt}\left( \frac{20e^{-rt}}{3r^2} + \frac{800e^{-rt}}{r} + \frac{20e^{-rt}t}{3r} + \mathrm{C} \right)[/tex]

[tex]S(t) = \frac{20t}{3r} + \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} e^{rt}[/tex]

[tex]t=0 \Rightarrow \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} = S_0 \Rightarrow \mathrm{C} = \frac{-20-2400r+3r^2 S_0}{3r^2} [/tex]

[tex]S(t)=\frac{20}{3r^2} - \frac{20e^{rt}}{3r^2} + \frac{800}{r} - \frac{800e^{rt}}{r} + S_0 e^{rt} + \frac{20t}{3r}[/tex]

[tex]S(t)=0 \Rightarrow t \approx - 131 \mbox{ months} [/tex]

which is WRONG!

Any help is highly appreciated.
 
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  • #2
thiago_j said:
"A recent college graduate borrows $100,000 at an interest rate of 9% to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800(1 + t/120), where t is the number of months since the loan was made. Assuming that this payment schedule can be maintained, when will the loan be fully paid?"

Please, help me find where I made a mistake. Here's what I've got:

[tex]S_0 = \$ 100,000 [/tex]

[tex]r = 0.09[/tex]

[tex]k(t) = \$ 800 \left( 1 + \frac{t}{120} \right) / \mbox{month}[/tex]

[tex]\frac{dS}{dt}=rS-k(t), \qquad S(0)=S_0[/tex]

[tex]\frac{dS}{dt}-rS=-k(t)[/tex]

[tex]\mu = \exp \left( -r \int dt \right) = e ^{-rt}[/tex]

[tex]S(t)=e^{rt}\int -800 \left( 1 + \frac{t}{120} \right) e ^{-rt} \: dt[/tex]

[tex]S(t)=e^{rt}\left( \frac{20e^{-rt}}{3r^2} + \frac{800e^{-rt}}{r} + \frac{20e^{-rt}t}{3r} + \mathrm{C} \right)[/tex]

[tex]S(t) = \frac{20t}{3r} + \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} e^{rt}[/tex]

[tex]t=0 \Rightarrow \frac{800}{r} + \frac{20}{3r^2} + \mathrm{C} = S_0 \Rightarrow \mathrm{C} = \frac{-20-2400r+3r^2 S_0}{3r^2} [/tex]

[tex]S(t)=\frac{20}{3r^2} - \frac{20e^{rt}}{3r^2} + \frac{800}{r} - \frac{800e^{rt}}{r} + S_0 e^{rt} + \frac{20t}{3r}[/tex]

[tex]S(t)=0 \Rightarrow t \approx - 131 \mbox{ months} [/tex]

which is WRONG!

Any help is highly appreciated.

I can't say I completely follow what you are doing, but it seems to me you have two different time scales going on and you may have lost track of the initial value a few equations into your analysis. If you were making no payments, the amount owed would be increasing every month. In the continuous interest limit you would have

[tex]S(t)=S_0 e^{rt/12}[/tex]

where r is the annual interest rate and t is in months. The payments were already expressed in terms of t in months and of course they reduce the value of S(t). So I think you need

[tex]\frac{dS}{dt}=rS/12 -k(t), \qquad S(0)=S_0[/tex]
 
  • #3
Your integration is all correct, so it is something small, OlderDan seems to be making sense to me.
 
  • #4
You're right. It's something pretty small. I should have divided r by 12, which ultimately gives the correct answer: t = 135.36 months.

Thanks
 

What is a First Order Differential Equation?

A First Order Differential Equation is a mathematical equation that relates an unknown function to its derivative. It involves only the first derivative of the function and can be solved to find the function itself.

What are some examples of First Order Differential Equations?

Some examples of First Order Differential Equations include the Newton's Law of Cooling, the Logistic Growth Model, and the Radioactive Decay Model. These equations are used to model various physical and real-world phenomena.

How do you solve a First Order Differential Equation?

The general method for solving a First Order Differential Equation involves separating the variables, integrating both sides, and then solving for the unknown function. Various techniques such as substitution, integration by parts, and separation of variables can be used to solve different types of First Order Differential Equations.

What is the significance of First Order Differential Equations in science?

First Order Differential Equations are used in many scientific fields such as physics, chemistry, biology, and engineering. They are essential in modeling and understanding complex systems and predicting how they will change over time.

Are there any real-world applications of First Order Differential Equations?

Yes, First Order Differential Equations have numerous real-world applications. They are used in predicting population growth, analyzing chemical reactions, and understanding the behavior of electrical circuits, among many others. They are also used in various fields of engineering, such as control systems and signal processing.

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