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Solutions to a Simple Differential Equation

  • Thread starter icefall5
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  • #1
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Homework Statement


Determine the values of r for which the given differential equation has solutions of the form [itex]y=t^r[/itex] for [itex]t > 0[/itex].


Homework Equations


[itex]t^2 y'' - 13ty' + 48y = 0[/itex]


The Attempt at a Solution


The program (online) has a thing that walks me through the question. It first had me find the second and first derivatives of [itex]y=t^r[/itex], which are [itex](r-1)*(r)*(t^{r-2})[/itex] and [itex]rt^{r-1}[/itex], respectively. It then tells me to plug those into the original equation, which gives me [itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r)[/itex]. I can apparently simplify that to [itex]r^2 - 14r + 48 = 0[/itex], but I have no idea how that works. I have not continued with the problem.
 
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Answers and Replies

  • #2
33,508
5,194

Homework Statement


Determine the values of r for which the given differential equation has solutions of the form [itex]y=t^r[/itex] for [itex]t > 0[/itex].


Homework Equations


[itex]t^2 y'' - 13ty' + 48y = 0[/itex]


The Attempt at a Solution


The program (online) has a thing that walks me through the question. It first had me find the second and first derivatives of [itex]y=t^r[/itex], which are [itex](r-1)*(r)*(t^{r-2})[/itex] and [itex]rt^{r-1}[/itex], respectively. It then tells me to plug those into the original equation, which gives me [itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r)[/itex]. I can apparently simplify that to [itex]r^2 - 14r + 49 = 0[/itex], but I have no idea how that works. I have not continued with the problem.
Continuing from where you left off:
##(t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0##
So ##(r^2 - r -13r + 48)t^r = 0##
Or ##(r^2 - 14r + 48)t^r = 0##

What must be happen for the equation above to be true?
 
  • #3
CAF123
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I get to a simplified form [itex] t^r( r^2 -14r +48) = 0. [/itex]
I think you have mistyped 48 as 49?
 
  • #4
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Continuing from where you left off:
##(t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0##
So ##(r^2 - r -13r + 48)t^r = 0##
Or ##(r^2 - 14r + 48)t^r = 0##

What must be happen for the equation above to be true?
I'm trying to simplify it (that's what's getting me), and I can simplify it to [itex](t^r)(r-1) - 13 + 48[/itex]. I'm clearly doing something wrong, but I don't know what.
I get to a simplified form [itex] t^r( r^2 -14r +48) = 0. [/itex]
I think you have mistyped 48 as 49?
It is 48, my apologies. Thanks!
 
  • #5
CAF123
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Know that, for example, [itex] t^2(r^2t^{r-2}) = r^2t^r. [/itex]
Do you see whats happening here?
 
  • #6
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Know that, for example, [itex] t^2(r^2t^{r-2}) = r^2t^r. [/itex]
Do you see whats happening here?
Ah, that makes sense. Okay, using that, I now follow these steps to simplify:

[itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0[/itex]
[itex](t^r)(r-1)(r) - (13)(r) + (48)(t^r) = 0[/itex]
[itex](t^{r}r^{2}-t^{r}r) - 13r + 48t^r = 0[/itex]
[itex]t^{r}(r^{2}-r)-13r-48t^{r} = 0[/itex]
 
  • #7
CAF123
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Close, [itex] -13trt^{r-1} = -13rt^r, [/itex] and now use [itex] t^r [/itex] as a common factor.
 
  • #8
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Close, [itex] -13trt^{r-1} = -13rt^r, [/itex] and now use [itex] t^r [/itex] as a common factor.
Ah, okay, for some reason I canceled it out rather than writing [itex]t^r[/itex]. Copy-pasting my work from above and editing appropriately now gives me the below output, which should be correct.

[itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0[/itex]
[itex](t^r)(r-1)(r) - (13)(r)(t^r) + (48)(t^r) = 0[/itex]
[itex](t^{r}r^{2}-t^{r}r) - 13rt^{r} + 48t^r = 0[/itex]
[itex]t^{r}(-r^2 - r - 13r + 48) = 0[/itex]
[itex]t^{r}(-r^2 - 14r + 48) = 0[/itex]

With that, I need to solve for t, correct? Then I'll be able to easily find the values of r for which [itex]t > 0[/itex].
 
  • #9
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5,194
I'm trying to simplify it (that's what's getting me), and I can simplify it to [itex](t^r)(r-1) - 13 + 48[/itex]. I'm clearly doing something wrong, but I don't know what.
What you show is incorrect, and you have lost sight of the fact that you're dealing with an equation. You are not simplifying the equation I showed, which was
## (r^2 - 14r + 48)t^r = 0##

The goal is to solve for r, not t.

This is not a hard problem. You've already done all of the hard work.

Under what circumstances (i.e., for what r) does ## (r^2 - 14r + 48)t^r = 0##
 
  • #10
33,508
5,194
Ah, okay, for some reason I canceled it out rather than writing [itex]t^r[/itex]. Copy-pasting my work from above and editing appropriately now gives me the below output, which should be correct.

[itex](t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r) = 0[/itex]
[itex](t^r)(r-1)(r) - (13)(r)(t^r) + (48)(t^r) = 0[/itex]
[itex](t^{r}r^{2}-t^{r}r) - 13rt^{r} + 48t^r = 0[/itex]
[itex]t^{r}(-r^2 - r - 13r + 48) = 0[/itex]
[itex]t^{r}(-r^2 - 14r + 48) = 0[/itex]

With that, I need to solve for t, correct? Then I'll be able to easily find the values of r for which [itex]t > 0[/itex].
You have several errors in your work above, including a couple of sign errors.
 
  • #11
CAF123
Gold Member
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A stray minus cropped in the [itex] r^2 [/itex] term, but other than that you now have to solve for r. [itex] t^r ≠ 0 [/itex] (the graph is similar to an exponential). So the other term must equal 0.
 
  • #12
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0
A stray minus cropped in the [itex] r^2 [/itex] term, but other than that you now have to solve for r. [itex] t^r ≠ 0 [/itex] (the graph is similar to an exponential). So the other term must equal 0.
Okay, I apologize for being an idiot. I was looking at the [itex]t > 0[/itex] in the problem and thinking that's what I was solving for.

The corrected final equation should be [itex]t^{r}(r^{2}-14r+48) = 0[/itex], as Mark44 said. Using that formula, r = 6 and r = 8. Submitted those and they are, in fact, correct.

After reading through everything again, I apologize for not getting how to do it -- it should've been simple. Thank you all for your help, I really appreciate it!
 

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