Turn a Solved DE into a Taylor Series

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Many of you have probably used the book Differential Equations by Lomen & Lovelock.
For my class I'm working on Example 2, Page 153.

You don't need to see the book, though, to help me out. It's a four-part problem and I'm on the last step not knowing where to take it.

In Part B, we created a DE:

dP/dt = (397/4169) P + (6580236/4169)

and solved it:

P(t) = (397/8338) t^2 + (6580236/4169) t + (1225953/4169)

Now in Part D they ask:
"Expand the explicit solution you obtained in Part B in a Taylor Series about the origin."

I'm not so sure I've ever done anything like this, so I'm not clear on what steps I should take. I believe my teacher said I should find a similar Taylor Series and manipulate it to fit my needs but that still doesn't get me very far.

Absolutely any insight is appreciated, hope you can help me!

Thanks,
Eric
 

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  • #2
HallsofIvy
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If your differential equation really is
dP/dt = (397/4169) P + (6580236/4169)
then dP/dt = (397/4169) P + (6580236/4169)
is NOT the solution. Your P(t) satisfies dP/dt = (397/4169) t + (6580236/4169).

The solution to dP/dt = (397/4169) P + (6580236/4169)
will involve an exponential function.
 
  • #3
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I think I'm going to have to look at it in a different light. I could've sworn I put it correctly into maple, that's what it threw back at me.

Now that I try again, I get P(t) = -315680/19+exp((19/200)*t)*_C1, which I have checked by hand. I'm not entirely sure it's right though, I solved for C using a point from my original data table and got funny results when trying to use it.

I'll try again tomorrow. But now that I have an e in it, I believe I'll just be modifying the taylor series for e.
 
  • #4
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Okay, I've figured out the problem. The one I posted was from a different part of the question.

I believe my correct equation is
[tex]P(t)=e^{.095t}16543 - 16571[/tex]

Now I know that E is
[tex]e^{x} = \Sigma\frac{x^{n}}{n!}[/tex]

So, correct me if I'm wrong, but do I manipulate the series as so to match my equation?

[tex]\Sigma[\frac{(0.095t)^n}{n!}16543-16571][/tex]

To be proper, all the summations here are from n=0 to [tex]\infty[/tex]

In the question they ask me to relate it to the solution I got in part A, which is similar to what I originally posted:
[tex]P(t) = a t^2 + b t + c[/tex]

I don't think how I have it set up results in an answer in that form. Did I turn it into a taylor series properly?
 
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  • #5
HallsofIvy
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Okay, I've figured out the problem. The one I posted was from a different part of the question.

I believe my correct equation is
[tex]P(t)=e^{.095t}16543 - 16571[/tex]

Now I know that E is
[tex]e^{x} = \Sigma\frac{x^{n}}{n!}[/tex]

So, correct me if I'm wrong, but do I manipulate the series as so to match my equation?

[tex]\Sigma[\frac{(0.095t)^n}{n!}16543-16571][/tex]
No, the "-16571" is NOT part of the exponential and is NOT summed an infinite number of times!

To be proper, all the summations here are from n=0 to [tex]\infty[/tex]

In the question they ask me to relate it to the solution I got in part A, which is similar to what I originally posted:
[tex]P(t) = a t^2 + b t + c[/tex]

I don't think how I have it set up results in an answer in that form. Did I turn it into a taylor series properly?
And, again, that quadratic is WRONG.
 
  • #6
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Well stay with me here...

I was unsure and that's what I was asking about, now as I understand it my new equation should be:
[tex]P(t) = (16543)\Sigma[\frac{(0.095t)^n}{n!}]-16571[/tex]

True, the [tex]P(t) = a t^2 + b t + c[/tex] is incorrect for the original DE I gave you, but it is correct for a different part of the problem. This step is relating the quadratic to our new equation which we're making with summations.


I think I did this right, to relate them, I calculated out the summation to three terms and plugged them in to get:
[tex]P(t) = 74.7t^2+1571.6t-28[/tex]

which I can relate to my quadratic using similarities.
 
  • #7
HallsofIvy
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Well stay with me here...

I was unsure and that's what I was asking about, now as I understand it my new equation should be:
[tex]P(t) = (16543)\Sigma[\frac{(0.095t)^n}{n!}]-16571[/tex]
Good! This is now correct.

True, the [tex]P(t) = a t^2 + b t + c[/tex] is incorrect for the original DE I gave you, but it is correct for a different part of the problem. This step is relating the quadratic to our new equation which we're making with summations.


I think I did this right, to relate them, I calculated out the summation to three terms and plugged them in to get:
[tex]P(t) = 74.7t^2+1571.6t-28[/tex]

which I can relate to my quadratic using similarities.
Okay, I didn't do the calculations myself but assuming the arithmetic is correct, this is the "2nd order Taylor Polynomial" for your function.
 

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