Differential Equation Model


by roam
Tags: differential, equation, model
roam
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#1
Apr3-12, 04:57 AM
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1. The problem statement, all variables and given/known data

Here's a model for the balance owed on a loan with the following conditions:

* Interest accumulated on the loan at a rate of 5.24% per year

* The amount owed at the beginning of the loan was $20,000.

* No payments were made on the loan for the first two years

* After two years, the loan was paid off at $3,000 per year.

The model is given by the equations:

[itex]\frac{dL}{dt} = 0.0524 \ L, \ \ \ 0<t<2[/itex]

[itex]\frac{dL}{dt} = 0.0524 \ L \ - \ 3 , \ \ t>2[/itex]

(L is in thousands)

Using a direction field work out how large repayments should be if the loan is to be paid back in exactly 12 years (i.e., with exactly 10 years of repayments after the first two years with no repayments).

3. The attempt at a solution

Here is the direction field I made for this model



And the solution for the initial value L(0)=20 is



The root of this solution occurs at t=11.39, so 11 years is how long it will take to pay back the loan.

But how can we work out how large repayments should be if the loan is to be paid back in 12 years?

Any help is greatly appreciated.
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Ray Vickson
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#2
Apr3-12, 12:42 PM
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Change the DE to dL/dt = 0.0524L - R for t > 2, and solve it with R as a symbolic parameter. Then find R that makes L(12) = 0.

RGV
roam
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Apr3-12, 04:11 PM
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Quote Quote by Ray Vickson View Post
Change the DE to dL/dt = 0.0524L - R for t > 2, and solve it with R as a symbolic parameter. Then find R that makes L(12) = 0.

RGV
So, [itex]L(12)= 0.0524 \times (12) - R = 0.6288 - R = 0 \implies R = 0.6288[/itex]

Is this what you meant?

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Apr3-12, 04:30 PM
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Differential Equation Model


Quote Quote by roam View Post
So, [itex]L(12)= 0.0524 \times (12) - R = 0.6288 - R = 0 \implies R = 0.6288[/itex]

Is this what you meant?
No, it is not what I said and not what I meant. Go back and read what I wrote. You need to solve the DE for t > 2.

RGV
roam
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#5
Apr3-12, 04:56 PM
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Quote Quote by Ray Vickson View Post
No, it is not what I said and not what I meant. Go back and read what I wrote. You need to solve the DE for t > 2.

RGV
I didn't quite understand what you meant by treating R as a "symbolic parameter". I already found the value of R that gives 0. So we can write the equation as

dL/dt = 0.0524 L - 0.6288

I'm not sure how this helps us. Did you mean I have to first solve the DE using separation of variables while ignoring R? I appreciate it if you could maybe explain that a bit more clearly.
epenguin
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#6
Apr3-12, 06:14 PM
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It looks like you are using a phase plane plotter as well as the direction field. You can use it or the direction field to plot backwards as well as forwards in time! (And it looks like that is what your plotter is doing since your starting point is in the middle!)
roam
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#7
Apr3-12, 06:26 PM
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Yes I've used the Matlab add-on called "dfield", my plot shows the initial condition y(0)=20. But how do can we use this plot to figure out how large repayments should be if the loan is to be paid back in exactly 12 years?
Ray Vickson
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Apr3-12, 07:19 PM
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I don't have access to Matlab, so I cannot offer technical advice. But if I were doing the question I would avoid the use of a phase plot and would, instead, just write down the formula for the DE solution; it would be a formula that has both R and t in it: L = f(t,R). Then I would set L=0 at t=12 and solve the equation to find R.

I guess if you want to use phase plot methods you could start by guessing a value of R, then make the phase plot for that R value and trace out the solution. If L=0 occurs before t=12 our guessed value if R is too large, so we should decrease it a bit and start over. If L=0 occurs after t=12, our guessed R is too small, so we should increase it a bit and start over. This procedure would be horrible and would take forever, and that is why I would not use it unless I had hours of spare time and nothing better to do.

RGV
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Apr3-12, 07:22 PM
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Your plot is actually predicting backwards from that t(0) - predicting your what your debt would have been over the previous 30+ years if your first eq. had been applicable over that time!

Now you know where your starting t is for the 'retrodiction'. t = 12. It should be obvious that at your present R = 3 the slopefield will not take you back from (12, 0) to (0, 20) i.e. L = 20 but to something higher. If not obvious make a few trials and you'll soon see. So as you cannot get back to 20 at the right time with that R try different R till it comes right!

You can as RV said calculate it mathematically, which is not very hard depending on what math you know. But using the plotter is not even math, it's common sense.
roam
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Apr3-12, 07:24 PM
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Quote Quote by Ray Vickson View Post
I don't have access to Matlab, so I cannot offer technical advice. But if I were doing the question I would avoid the use of a phase plot and would, instead, just write down the formula for the DE solution; it would be a formula that has both R and t in it: L = f(t,R). Then I would set L=0 at t=12 and solve the equation to find R.

I guess if you want to use phase plot methods you could start by guessing a value of R, then make the phase plot for that R value and trace out the solution. If L=0 occurs before t=12 our guessed value if R is too large, so we should decrease it a bit and start over. If L=0 occurs after t=12, our guessed R is too small, so we should increase it a bit and start over. This procedure would be horrible and would take forever, and that is why I would not use it unless I had hours of spare time and nothing better to do.

RGV
I agree, that would take a lot of time. So could you explain to me how to find the right value mathematically without using the phase plots? (I didn't quite get your first post)
epenguin
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Apr3-12, 07:26 PM
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Quote Quote by roam View Post
I agree, that would take a lot of time. So could you explain to me how to find the right value without using the phase plots? (I didn't quite get your first post)
How did you get the continuous curve that is in your pic?
roam
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Apr3-12, 07:43 PM
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Quote Quote by epenguin View Post
How did you get the continuous curve that is in your pic?
I just went to the dfield set-up window and typed in the initial condition x=20 when t=0, and it gave me the curve.
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Apr3-12, 07:49 PM
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Quote Quote by roam View Post
I just went to the dfield set-up window and typed in the initial condition x=20 when t=0, and it gave me the curve.
It sounds like you have got a programme or calculator that (like most such) doesn't do only d-fields but plots phase planes or paths as well (fancy name for 'solves 1st order d.e.'s in 2 variables').
You have all you need.

(I have to go now.)
Ray Vickson
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Apr3-12, 07:52 PM
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Quote Quote by roam View Post
I agree, that would take a lot of time. So could you explain to me how to find the right value mathematically without using the phase plots? (I didn't quite get your first post)
Since you titled your thread "differential equation model" I assumed you knew something about differential equations. Is that assumption wrong?

I really need more information from you in order to _guide_ you while avoiding doing your homework for you. What is the course? What is your background? How much calculus have you had? etc.

RGV
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#15
Apr3-12, 08:01 PM
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Quote Quote by Ray Vickson View Post
Since you titled your thread "differential equation model" I assumed you knew something about differential equations. Is that assumption wrong?

I really need more information from you in order to _guide_ you while avoiding doing your homework for you. What is the course? What is your background? How much calculus have you had? etc.

RGV
I have done basic calculus and some linear algebra. But this is my first course in differential equations. So far I've learned how to solve DEs using the method of separation of variables and integrating factor.
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Apr3-12, 08:08 PM
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Quote Quote by roam View Post
I have done basic calculus and some linear algebra. But this is my first course in differential equations. So far I've learned how to solve DEs using the method of separation of variables and integrating factor.
Well, that is plenty enough to solve the simple DE dL/dt = c*L - R with constants c and R. Just apply what you have been taught.

It may be that you are comfortable with solving the DE dL/dt = c*L but uncomfortable with the DE that has the extra '-R' on the right. Is that the case? If so, look instead at the DE for M = L - p, where p is some constant. You ought to be able to figure out what p value to choose in order to get rid of the constants on the right and have just dM/dt = c*M.

Alternatively, you can use an integrating factor.

RGV


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