calculus and ode's

**simfonija** · October 20th 2006, 03:56 AM

please help me with this. It's urgent and I would appreciate any help. Thanks

**dan** · October 20th 2006, 04:37 AM

you need help with the whole test?

**simfonija** · October 20th 2006, 04:44 AM

Yes, I don't have any notes and have real problems to solve it. Thanks in advance

**dan** · October 20th 2006, 04:46 AM

I need to go to work now but if no one esle helps I can try later.
though I might need help on some of it

dan

**Soroban** · October 20th 2006, 04:52 AM

Hello, simfonija!

You need help with the entire exam?
Should we be helping you?

Having said that, here's #1 . . .

1) Find the interval of convergence of the power series: . $\sum^{\infty}_{n=0}\frac{(1 - \frac{x}{2})^n}{\sqrt{n+1}}$

Ratio Test

$r \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(1 - \frac{x}{2})^{n+1}}{\sqrt{n+2}} \cdot \frac{\sqrt{n+1}}{(1 - \frac{x}{2})^n} \;=\;\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)$

$R\;=\;\lim_{n\to\infty}|r|\;=\;\lim_{n\to\infty}\l eft|\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)\right| \;= \;\left|1 - \frac{x}{2}\right|$ **

For convergence, $R < 1$

So we have: . . . . $\left|1 - \frac{x}{2}\right| \; < \; 1$

Then: . . . . . $-1 \; < \; 1 - \frac{x}{2} \; < \; 1$

Subtract 1: . . $-2 \; < \; -\frac{x}{2} \; < \; 0$

Multiply by -2: . . $4 \; > \; x \; > \; 0$

Therefore: . . . . $\boxed{0 \: < \: x \: < \: 4}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
We have: . $\lim_{n\to\infty}\sqrt{\frac{n+1}{n+2}}$

Divide top and bottom by $n:\;\;\sqrt{\frac{\frac{n}{n} + \frac{1}{n}}{\frac{n}{n} + \frac{2}{n}}} \;=\;\sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}}$

Then: . $\lim_{n\to\infty} \sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}} \;= \;\sqrt{\frac{1 + 0}{1 + 0}} \;=\;\sqrt{1} \;= \;1$

**simfonija** · October 20th 2006, 05:27 AM

Hi and thanks. I know I am asking a lot but with no notes I am lost. And again please help with the rest

**topsquark** · October 20th 2006, 05:46 AM

Originally Posted by simfonija

Hi and thanks. I know I am asking a lot but with no notes I am lost. And again please help with the rest

Please forgive me, but I am curious. What happened to your notes?

-Dan

**simfonija** · October 20th 2006, 06:35 AM

Ok, are you playing a teacher and I am bad student??? I was away for more than month (serious family reason) and hoping to manage to pass exam somehow. This is assignment which gives me 3% of total mark. On Monday I ll get copy of the notes. Just help if you can (no provocation please), if can't I also understand... Thanks

**topsquark** · October 20th 2006, 07:00 AM

Originally Posted by simfonija

Ok, are you playing a teacher and I am bad student??? I was away for more than month (serious family reason) and hoping to manage to pass exam somehow. This is assignment which gives me 3% of total mark. On Monday I ll get copy of the notes. Just help if you can (no provocation please), if can't I also understand... Thanks

My apologies. I was not intending to attack you, I was simply asking a question.

-Dan

**ThePerfectHacker** · October 20th 2006, 07:32 AM

I have something to add to Soroban's post.
He did not find the entire interval of convernge.
The endpoints are,
$x=0,x=4$
The two respective series we obtain are,
$\sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}}$
This can the zeta function $\zeta(1/2)$ , alternatively you can use the p-series test for exponent 1/2.
And also, the function,
$f(x)=\frac{1}{\sqrt(x+1)}$ is a positive continous decreasing function and the improper integral:
$\int_0^{\infty} \frac{dx}{\sqrt{x+1}}$ diverges. Thus, $x\not =0$

Checking the point at $x=4$ we have,
$\sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}}$
We have that $\lim_{n\to \infty}a_k=0$
The series is strictly decreasing and alternating. Thus, by Leibniz's test it is convergent.
$0<x\leq 4$

**simfonija** · October 20th 2006, 02:01 PM

Does anybody have idea how to solve 2 and 3 please

**AfterShock** · October 20th 2006, 05:21 PM

You're not going to learn how to do them if you have someone do ALL of them. Obviously this is a homework assignment that they expected you should be able to do. Did you even attempt any of these?

**simfonija** · October 20th 2006, 06:14 PM

Hi
Task 2 I have no idea how to solve. Task number 3 I tried and I did some calculation but 3 (a) I don’t know why 6^1/2=2(1+1/2)^1/2 is better than the other. For 4 I also don’t have example of such diff equations. I am not so bad in maths but as I said this assignment is for Monday and on Monday I will copy the notes from the colleague.

**ThePerfectHacker** · October 21st 2006, 06:20 PM

#4)
Solve, for $x>0$ .
$x^2y''-xy'+y=0$ .
Since, this is equivalent to,
$y''-\frac{1}{x}y'+\frac{1}{x^2}y=0$ where $p(x),q(x)$ are continous on some open interval there exists a basis of dimension two. Hence two linearly independant solutions.
We are given that $y_1=x$ is a solution.
To find, the second linearly independent solution we use reduction of order (which leads to):
$y_2=y_1\int \frac{W(y_1,y_2)}{y_1^2} dx$
The Wronskian (we can remove the constant multiplier) by Abel's theorem is,
$\mbox{exp} \left( -\int p(x) dx \right)=e^{\ln |x|}=x$
Thus,
$y_2=x\int \frac{x}{x^2}dx=x\ln x$
We see that this is a solution to this differencial equation which is linearly independent to $y_1=x$
Thus, the set of all solutions can be expressed as (general solution),
$C_1 x+C_2 x\ln x$
Congratulation we solved the homogenous equation.
----
Now the non-homogenous equation.
For the purposes of the solution we need to find the Wronskian between these two solutions which is,
$W(y_1,y_2)=\left| \begin{array}{cc} x&x\ln x\\1&1+\ln x \end{array} \right|=x$

Now we use Variation of Parameters.
The formula is,
$-y_1\int \frac{y_2 g(x)}{W(y_1,y_2)}dx+y_2\int \frac{y_1 g(x)}{W(y_1,y_2)}dx$
Thus,
$-x\int \frac{x\ln x\cdot x^{5/2}}{x} dx+x\ln x\int \frac{x\cdot x^{5/2}}{x} dx$
Thus,
$-x\int x^{5/2}\ln x dx+x\ln x\int x^{5/2} dx$
Integrate use by parts on first integral,
$-x\left( \frac{2}{7}\ln x x^{7/2} - \frac{4}{49} x^{7/2} \right) +x\ln x \left( \frac{2}{7}x^{7/2} \right)$
Thus,
$\frac{4}{49}x^{9/2}-\frac{2}{7}x^{9/2}\ln x+\frac{2}{7}x^{9/2}\ln x$
Thus,
$\frac{4}{49}x^{9/2}$
Thus, all solutions can be expressed as,
$C_1 x+C_2 x\ln x+\frac{4}{49}x^{9/2}$

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