Hi, I was wondering if anyone could help me with this problem. The differential equation y'=4y/x has a power function solution y=Ax^n if y(3)=162, determine the values of n and A. Thanks!

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- Oct 21st 2012, 10:34 PMchyksparks333Differential Equations?
Hi, I was wondering if anyone could help me with this problem. The differential equation y'=4y/x has a power function solution y=Ax^n if y(3)=162, determine the values of n and A. Thanks!

- Oct 21st 2012, 10:45 PMMarkFLRe: Differential Equations?
The ODE is separable...can you write it in this form?

- Oct 21st 2012, 10:49 PMchyksparks333Re: Differential Equations?
i can honestly say, i don't know what you're talking about haha i'm sorry i just started calc a month ago...

- Oct 21st 2012, 10:56 PMMarkFLRe: Differential Equations?
I'm sorry, ODE is short for ordinary differential equation. I assumed you were beginning a course in ordinary differential equations.

When you separate the variables, you typically get everything involving the dependent variable on the left and the independent variable on the right. For this ODE, you would then have:

$\displaystyle \frac{1}{y}\.dy=\frac{4}{x}\,dx$

Now, integrate...what do you get? - Oct 21st 2012, 11:02 PMchyksparks333Re: Differential Equations?
I'm still a little confused with all this new terminology but are you saying that on the left it would be 1/y (162) = 4/x (3)...I might be completely wrong...sorry

- Oct 21st 2012, 11:13 PMMarkFLRe: Differential Equations?
When you integrate, you will have:

$\displaystyle \int\frac{1}{y}\,dy=4\int\frac{1}{x}\,dx$

I am assuming you have been taught $\displaystyle \int\frac{1}{u}\,du=\ln|u|+C$. You need only write the constant of integration on one side, and it is typically written on the right side with the independent variable. It can be thought of as the composite constant from both integrations.

See if you can write the next step, and from there we will use the properties of logs to get the desired form. - Oct 21st 2012, 11:15 PMchyksparks333Re: Differential Equations?
I have not been taught that form before. is that the only way to attack this problem? because i feel a little lost at the moment.

- Oct 21st 2012, 11:26 PMMarkFLRe: Differential Equations?
Hmm...if you have not seen that rule of integration, then perhaps you are to just take the given solution:

(1) $\displaystyle y=Ax^n$

and then differentiate it with respect to $\displaystyle x$ and write the resulting differential equation in the given form. Using this and the point on the solution curve (3,162), you will then have two equations and two unknowns from which you can determine $\displaystyle A$ and $\displaystyle n$.

What do you get when you differentiate (1) with respect to $\displaystyle x$?

I apologize for the confusion, I assumed you were to solve the ODE and show that the solution is of the given form. - Oct 21st 2012, 11:30 PMchyksparks333Re: Differential Equations?
Without any numbers plugged in, i assume it would be y= Anx^n-1 but if you put in (3,162), would it be, 162=A(3n)^n-1

- Oct 22nd 2012, 12:03 AMSironRe: Differential Equations?
First, compute the derivative

$\displaystyle y = Ax^{n} \Rightarrow y' = Anx^{n-1}$

Substitute $\displaystyle y$ and $\displaystyle y'$ in the differential equation:

$\displaystyle Anx^{n-1} = \frac{4Ax^n}{x}$

$\displaystyle \Leftrightarrow n=4$

At this point we can use that $\displaystyle y(3)=162$ thus

$\displaystyle y(3)=162 = 81A \Leftrightarrow A = 2$

Therefore the solution is:

$\displaystyle y=2x^4$ - Oct 22nd 2012, 12:20 AMMarkFLRe: Differential Equations?
I guess I was taking too long! LOL!

- Oct 24th 2012, 05:42 PMtopsquarkRe: Differential Equations?
@ chyksparks333

Since you have decided to not allow PMs we have to do this in public. You reported this post and said that there was something in the thread that was "not cool." I have reviewed the thread and I don't see what you could be reacting to. Please let me know as soon as you can.

-Dan - Oct 24th 2012, 06:09 PMMarkFLRe: Differential Equations?
I hope it wasn't my previous comment...I was only trying to be funny. :)

- Oct 27th 2012, 03:25 PMtopsquarkRe: Differential Equations?
- Oct 27th 2012, 04:24 PMHallsofIvyRe: Differential Equations?
I'm a bit late on this but note that the problem does NOT ask you to

**solve**the equation. You are**given**that the solutions is $\displaystyle y= Ax^n$ for some A and n and you are asked to find A and n. Siron's method- find the derivative of y, then put that and y into the equation, is the correct idea.