Solutions to a Simple Differential Equation

[icefall5]

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.

 

[Mark44]

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?

 

[CAF123]

I get to a simplified form $t^r( r^2 -14r +48) = 0.$
I think you have mistyped 48 as 49?

 

[icefall5]

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.

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!

 

[CAF123]

Know that, for example, $t^2(r^2t^{r-2}) = r^2t^r.$
Do you see what's happening here?

 

[icefall5]

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:

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

 

[CAF123]

Close, $-13trt^{r-1} = -13rt^r,$ and now use $t^r$ as a common factor.

 

[icefall5]

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 $t > 0$.

 

[Mark44]

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 $(r^2 - 14r + 48)t^r = 0$

 

[Mark44]

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.

 

[CAF123]

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.

 

[icefall5]

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!