MATH1251 Questions HELP (2 Viewers)

Paradoxica · Aug 20, 2016

1008 said:
I've got a couple of ODEs too.

1)Use the substitution to solve:

$xyy''=yy'+x(y')^2;u=y'/y \text{ or }u=\ln y$

$\\\text{2) Suppose that }\alpha(x)\text{ is a solution to the equation }\\ \frac{\mathrm{d^2} u}{\mathrm{d} x^2} +b(x) \frac{\mathrm{d} u}{\mathrm{d} x} +c(x)u=0\\\text{(a) Use the substitution }u(x)=\alpha(x)v(x)\text{ to write down a first order equation for } \frac{\mathrm{d} v}{\mathrm{d} x}\\\text{What method can you see to solve this equation?}\\\text{(b) Verify that }u(x)=x^3\text{ is one of the solutions to the differential equation }\\x^2\frac{\mathrm{d^2} u}{\mathrm{d} x^2}-4x\frac{\mathrm{d} u}{\mathrm{d} x}+6u=0\\\text{Hence use the technique of (a) to find the general solution.}$

I got it...

$$\noindent From the question, we have:$ \\\\ \alpha'' + b\alpha' + c\alpha = 0 \\\\ u'' = (\alpha v)'' = u''\alpha + 2\alpha'v' + av'' \\\\ u' = (\alpha v)' = \alpha'v + \alpha v' \\\\ $Substituting this stuff in place of $u$ in the differential equation gives us:$ \\\\ \alpha''v + 2\alpha'v' + \alpha v'' + b \alpha' v + b\alpha v' + c\alpha v = 0 \\\\ $Collect the terms relevant to the original differential equation I wrote above, to obtain: $\\\\ (\alpha'' + b \alpha' + c\alpha)v + 2\alpha'v' + \alpha v'' + b\alpha v' = 0 \\\\ $Which vanishes, leaving us with:$ \\\\ 2\alpha'v' + \alpha v'' + b\alpha v' = 0 \\\\ $Divide by $\alpha v'$ to obtain:$ \\\\ 2 \frac{\alpha'}{\alpha} + \frac{v''}{v'} + b = 0 \\\\ $So then we get at last:$ \\\\ 2\log{\alpha} + \log{v'} = \int b(x) \text{d}x \\\\$And then the solution all boils down to evaluating the integral of $b(x).$

InteGrand · Aug 20, 2016

leehuan said:
Also to the people actually in the course, I'm just helping advertise the PASS classes that they are running. They are worth going imo though.

This question seems pretty tedious to write out (as in, both the qn and ans). I was gonna work my way up by finding m₀=f(t) but then I realised that if I did that then I'd be solving an endless stream of first order linear ODEs without guaranteeing I get a pattern. Is there a way to force the pattern out?

$\noindent I assume you got a) out (it's just using the principle of rate of inflow minus rate of outflow).$

$\noindent Now for b), for simplicity, let $a=0.03$. We prove by induction that $m_k = \frac{50a^k}{k!}t^k e^{-at}$ for all $k=0,1,\ldots, n$. The result holds for $k=0$, because we have $m_0 ^\prime = -am_0$, $m_0 (0) = 50$, which as we know has solution $m_0 (t) = m_0 (0) e^{-at} = 50e^{-at}$, which fits the proposition for $k=0$.$

$\noindent Now suppose the proposition holds for some particular integer $k$ with $0\leq k < n$, i.e. that $m_k (t) = \frac{50a^k}{k!}t^k e^{-at}$. We show that $m_{k+1} (t) = \frac{50a^{k+1}}{\left(k+1\right)!}t^{k+1} e^{-at}$. For simplicity, denote $y\equiv m_{k+1}$. Note $y(0) = 0$ (since the tanks after the 0-th one have no salt in them at the start).$

$\noindent From part a) and the inductive hypothesis, we have$

$\begin{align*}y^\prime &= am_{k} -ay \\ &= a \frac{50a^k}{k!}t^k e^{-at} -ay \\ \Rightarrow y^\prime + ay &= \frac{50a^{k+1}}{k!}t^k e^{-at}. \end{align*}$

$\noindent This is a linear first-order ODE. The integrating factor is $v = e^{at}$. So we get$

$\begin{align*}\left(vy\right)^\prime = \left(e^{at} y\right)^\prime &= v \frac{50a^{k+1}}{k!}t^k e^{-at} \\ &= e^{at}\cdot \frac{50a^{k+1}}{k!}t^k e^{-at} \\ &= \frac{50a^{k+1}}{k!}t^k \\ \Rightarrow e^{at} y &= \frac{50a^{k+1}}{k!} \cdot \frac{t^{k+1}}{k+1} + C \quad (\text{integrating both sides}) \\ \Rightarrow y &= \frac{50a^{k+1}}{\left(k+1\right)!} t^{k+1}e^{-at} + Ce^{-at}. \end{align*}$

$\noindent Recall that $y(0) = 0$, which implies that $C = 0$. Thus $y= \frac{50a^{k+1}}{(k+1)!}t^{k+1} e^{-at}$, and the proof is complete by induction. In particular, the result in the question is true by setting $k=n$.$

RenegadeMx · Aug 20, 2016

lol its scary for me that most of these ODES are harder than the ODES from the specific DE's course math2121/221, tho then again we did focus on more broader types of DE's

Paradoxica · Aug 20, 2016

RenegadeMx said:
lol its scary for me that most of these ODES are harder than the ODES from the specific DE's course math2121/221, tho then again we did focus on more broader types of DE's

Yeah I feel like most of these are just amenable to "that one trick you buried away infinity years ago in the back of your head" and all you have to do is spend like 10 minutes trying to pull it out of your bag of tricks.

Either that or just slick algebraic juggling.

1008 · Aug 22, 2016

leehuan said:
Also to the people actually in the course, I'm just helping advertise the PASS classes that they are running. They are worth going imo though.

This question seems pretty tedious to write out (as in, both the qn and ans). I was gonna work my way up by finding m₀=f(t) but then I realised that if I did that then I'd be solving an endless stream of first order linear ODEs without guaranteeing I get a pattern. Is there a way to force the pattern out?

Is this correct for part (b)?
IG, just regarding your approach, I thought of using induction, but if the question asks to "Show" and not "Prove" so can we still use induction (or is that a HSC thing only)?

InteGrand · Aug 22, 2016

1008 said:
Is this correct for part (b)?
IG, just regarding your approach, I thought of using induction, but if the question asks to "Show" and not "Prove" so can we still use induction (or is that a HSC thing only)?

The "there is a pattern" is basically proved using induction (just saying without proof that there is a pattern is kind of assuming the result). And induction is a valid proof method, so you should be allowed to use it (even in the HSC you can I think).

1008 · Aug 22, 2016

InteGrand said:
The "there is a pattern" is basically using induction. And induction is a valid proof method, so you should be allowed to use it.

Alright, thanks for clarifying that. I was just concerned that the question asked "show", not "prove"... And is this alright as a method?

EDIT: nvm haha, just read your above post

InteGrand · Aug 22, 2016

1008 said:
Alright, thanks for clarifying that. I was just concerned that the question asked "show", not "prove"... And is this alright as a method?

I don't think there's really any difference between 'show' and 'prove'. And I think you'd need to prove your claim about the pattern for k. This can be done using induction.

1008 · Aug 22, 2016

InteGrand said:
I don't think there's really any difference between 'show' and 'prove'. And I think you'd need to prove your claim about the pattern for k. This can be done using induction.

Yep, thanks

leehuan · Sep 15, 2016

$\text{a) Show that the Vandermonde determinant}\\ \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}=\prod_{i<j}(t_j-t_i)$

The only idea I had was to row reduce out the first column and then evaluate down it. And then I got stuck. Any suggestions?

$\begin{vmatrix}t_2-t_1&t_2^2-t_1^2&\dots&t_2^{n-1}-t_1^{n-1}\\ t_3-t_1&t_3^2-t_1^2&\dots &t_3^{n-1}-t_1^{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ t_n-t_1& t_n^2-t_1^2 & \dots & t_n^{n-1}-t_1^{n-1}\end{vmatrix}$

Paradoxica · Sep 15, 2016

leehuan said:
$\text{a) Show that the Vandermonde determinant}\\ \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}=\prod_{i<j}(t_j-t_i)$

The only idea I had was to row reduce out the first column and then evaluate down it. And then I got stuck. Any suggestions?

$\begin{vmatrix}t_2-t_1&t_2^2-t_1^2&\dots&t_2^{n-1}-t_1^{n-1}\\ t_3-t_1&t_3^2-t_1^2&\dots &t_3^{n-1}-t_1^{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ t_n-t_1& t_n^2-t_1^2 & \dots & t_n^{n-1}-t_1^{n-1}\end{vmatrix}$

subtract multiples of each row from the previous ones. remember that your "multiples" are not restricted to integers.

InteGrand · Sep 15, 2016

leehuan said:
$\text{a) Show that the Vandermonde determinant}\\ \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}=\prod_{i<j}(t_j-t_i)$

The only idea I had was to row reduce out the first column and then evaluate down it. And then I got stuck. Any suggestions?

$\begin{vmatrix}t_2-t_1&t_2^2-t_1^2&\dots&t_2^{n-1}-t_1^{n-1}\\ t_3-t_1&t_3^2-t_1^2&\dots &t_3^{n-1}-t_1^{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ t_n-t_1& t_n^2-t_1^2 & \dots & t_n^{n-1}-t_1^{n-1}\end{vmatrix}$

Another classic way is to use induction and view the determinant as a polynomial.

leehuan · Sep 15, 2016

Paradoxica said:
subtract multiples of each row from the previous ones. remember that your "multiples" are not restricted to integers.

$\text{True, but if I try the row operation }R_2\gets R_2-\frac{t_3-t_1}{t_2-t_1}R_1$

$A_{21}\text{ goes to }0\text{ but }A_{22}\text{ becomes messier.}$

What am I meant to use as my multiple?

Paradoxica · Sep 15, 2016

leehuan said:
$\text{True, but if I try the row operation }R_2\gets R_2-\frac{t_3-t_1}{t_2-t_1}R_1$

$A_{21}\text{ goes to }0\text{ but }A_{22}\text{ becomes messier.}$

What am I meant to use as my multiple?

x

leehuan · Sep 15, 2016

InteGrand said:
Another classic way is to use induction and view the determinant as a polynomial.

Oh? Determinant as a polynomial... mind linking me an article?

Paradoxica said:
x

...

Paradoxica · Sep 15, 2016

leehuan said:
Oh? Determinant as a polynomial... mind linking me an article?

...

Have you considered the factor theorem?

leehuan · Sep 15, 2016

Paradoxica said:
Have you considered the factor theorem?

Can't see how to use it

Paradoxica · Sep 15, 2016

leehuan said:
Can't see how to use it

t_i = t_j results in the matrix determinant vanishing.

InteGrand · Sep 15, 2016

leehuan said:
$\text{a) Show that the Vandermonde determinant}\\ \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}=\prod_{i<j}(t_j-t_i)$

The only idea I had was to row reduce out the first column and then evaluate down it. And then I got stuck. Any suggestions?

$\begin{vmatrix}t_2-t_1&t_2^2-t_1^2&\dots&t_2^{n-1}-t_1^{n-1}\\ t_3-t_1&t_3^2-t_1^2&\dots &t_3^{n-1}-t_1^{n-1}\\ \vdots & \vdots & \ddots & \vdots \\ t_n-t_1& t_n^2-t_1^2 & \dots & t_n^{n-1}-t_1^{n-1}\end{vmatrix}$

[This is not an original idea by me, this is a well-known problem with many solutions.]

$\noindent For positive integers $n$, let $V_{n} := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}$. Let $P_{n}(x) := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_{n-1} & t_{n-1}^2 & \dots & t_{n-1}^{n-1}\\ 1 & x & x^2 & \dots & x^{n-1}\end{vmatrix}$. It is easy to see this is a polynomial of degree $n-1$ in $x$ (e.g. expand along the bottom row to see this). So note that $V_{n} = P_{n} \left(t_{n}\right)$.$

$\noindent Note that if $t_i=t_j$ for some $i\neq j$, the determinant is $0$ and the formula holds. So assume now that all $t_j$ are distinct. It is easy to verify the formula if $n=1$ or $2$, so assume it holds for $n-1$. Now, observe that if we successively sub. $x=t_1,t_2,\ldots , t_{n-1}$ into $P_{n}(x) := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_{n-1} & t_{n-1}^2 & \dots & t_{n-1}^{n-1}\\ 1 & x & x^2 & \dots & x^{n-1}\end{vmatrix}$, the determinant vanishes (since we'll have two matching rows), so $P\left(t_j\right) = 0$ for $j=1,2,\ldots, n-1$. So we've found $n-1$ distinct roots of the polynomial, which as we noted was of degree $n-1$. So we have found all the roots and using the factor theorem can write $P_{n}(x) = C\prod _{j=1}^{n-1} \left(x-t_j\right)$. We just need to find the leading coefficient.$

$\noindent To find this, note that if we expand out the determinant along the last row (which is where the $x^{n-1}$ coefficient will come from), the coefficient of $x^{n-1}$ will just be the minor obtained by deleting the last row and column, which is just the same Vandermonde determinant but only going from $t_1$ to $t_{n-1}$ and from $1$ to power $n-2$ in each row. But this is just the Vandermonde determinant at one lower value of $n$, which by the inductive hypothesis is just $\prod _{1 \leq i < j \leq n-1} (t_j -t_i)$. So this is our $C$, so we have $P_{n}(x) =\prod _{1 \leq i < j \leq n-1} (t_j -t_i) \prod _{j=1}^{n-1} \left(x-t_j\right)$. So the determinant given to us, which is $P_n (t_{n})$, is $\prod _{1 \leq i < j \leq n-1} (t_j -t_i) \prod _{j=1}^{n-1} \left(t_{n}-t_j\right)$, which it is easy to see is the product formula we had to prove.$

leehuan · Sep 15, 2016

InteGrand said:
[This is not an original idea by me, this is a well-known problem with many solutions.]

$\noindent For positive integers $n$, let $V_{n} := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_n & t_n^2 & \dots & t_n^{n-1}\end{vmatrix}$. Let $P_{n}(x) := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_{n-1} & t_{n-1}^2 & \dots & t_{n-1}^{n-1}\\ 1 & x & x^2 & \dots & x^{n-1}\end{vmatrix}$. It is easy to see this is a polynomial of degree $n-1$ in $x$ (e.g. expand along the bottom row to see this). So note that $V_{n} = P_{n} \left(t_{n}\right)$.$

$\noindent Note that if $t_i=t_j$ for some $i\neq j$, the determinant is $0$ and the formula holds. So assume now that all $t_j$ are distinct. It is easy to verify the formula if $n=1$ or $2$, so assume it holds for $n-1$. Now, observe that if we successively sub. $x=t_1,t_2,\ldots , t_{n-1}$ into $P_{n}(x) := \begin{vmatrix}1&t_1&t_1^2&\dots&t_1^{n-1}\\ 1&t_2&t_2^2&\dots&t_2^{n-1}\\ 1&t_3&t_3^2&\dots&t_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots &\\ 1 & t_{n-1} & t_{n-1}^2 & \dots & t_{n-1}^{n-1}\\ 1 & x & x^2 & \dots & x^{n-1}\end{vmatrix}$, the determinant vanishes (since we'll have two matching rows), so $P\left(t_j\right) = 0$ for $j=1,2,\ldots, n-1$. So we've found $n-1$ distinct roots of the polynomial, which as we noted was of degree $n-1$. So we have found all the roots and using the factor theorem can write $P_{n}(x) = C\prod _{j=1}^{n} \left(x-t_j\right)$. We just need to find the leading coefficient.$

$\noindent To find this, note that if we expand out the determinant along the last row (which is where the $x^{n-1}$ coefficient will come from), the coefficient of $x^{n-1}$ will just be the minor obtained by deleting the last row and column, which is just the same Vandermonde determinant but only going from $t_1$ to $t_{n-1}$ and from $1$ to power $n-2$ in each row. But this is just the Vandermonde determinant at one lower value of $n$, which by the inductive hypothesis is just $\prod _{1 \leq i < j \leq n-2} (t_j -t_i)$. So this is our $C$, so we have $P_{n}(x) =\prod _{1 \leq i < j \leq n-2} (t_j -t_i) \prod _{j=1}^{n} \left(x-t_j\right)$. So the determinant given to us, which is $P_n (t_{n})$, is $\prod _{1 \leq i < j \leq n-2} (t_j -t_i) \prod _{j=1}^{n} \left(t_{n-1}-t_j\right)$, which it is easy to see is the product formula we had to prove.$

Amazing... Heaps of those ideas would not have crossed my mind tbh...

MATH1251 Questions HELP (2 Viewers)

-insert title here-

Well-Known Member

Kosovo is Serbian

-insert title here-

Active Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)