Solutions to a Simple Differential Equation

icefall5
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Homework Statement


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

Homework Equations


t^2 y'' - 13ty' + 48y = 0

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 y=t^r, which are (r-1)*(r)*(t^{r-2}) and rt^{r-1}, respectively. It then tells me to plug those into the original equation, which gives me (t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r). I can apparently simplify that to r^2 - 14r + 48 = 0, but I have no idea how that works. I have not continued with the problem.
 
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icefall5 said:

Homework Statement


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


Homework Equations


t^2 y'' - 13ty' + 48y = 0


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 y=t^r, which are (r-1)*(r)*(t^{r-2}) and rt^{r-1}, respectively. It then tells me to plug those into the original equation, which gives me (t^2)(r-1)(r)(t^{r-2}) - (13)(t)(r)(t^{r-1}) + (48)(t^r). I can apparently simplify that to r^2 - 14r + 49 = 0, 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?
 
I get to a simplified form t^r( r^2 -14r +48) = 0.
I think you have mistyped 48 as 49?
 
Mark44 said:
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 (t^r)(r-1) - 13 + 48. I'm clearly doing something wrong, but I don't know what.
CAF123 said:
I get to a simplified form t^r( r^2 -14r +48) = 0.
I think you have mistyped 48 as 49?
It is 48, my apologies. Thanks!
 
Know that, for example, t^2(r^2t^{r-2}) = r^2t^r.
Do you see what's happening here?
 
CAF123 said:
Know that, for example, t^2(r^2t^{r-2}) = r^2t^r.
Do you see what's happening here?
Ah, that makes sense. Okay, using that, I now follow these steps to simplify:

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

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

With that, I need to solve for t, correct? Then I'll be able to easily find the values of r for which t > 0.
 
icefall5 said:
I'm trying to simplify it (that's what's getting me), and I can simplify it to (t^r)(r-1) - 13 + 48. 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
icefall5 said:
Ah, okay, for some reason I canceled it out rather than writing t^r. Copy-pasting my work from above and editing appropriately now gives me the below output, which should be correct.

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

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

The corrected final equation should be t^{r}(r^{2}-14r+48) = 0, 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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