First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

RenegadeMx · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

u will see solving 2nd order ODES are piss easy

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

RenegadeMx said:
u will see solving 2nd order ODES are piss easy

Lol yeah they are. At least provided their coefficients stay constant, otherwise idk

Paradoxica · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Yeah I guess so. Makes sense

If you differentiate the second DE, you can solve simultaneously with the original DE to obtain a standard second order homogenous DE which you can then solve by substitution.

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
If you differentiate the second DE, you can solve simultaneously with the original DE to obtain a standard second order homogenous DE which you can then solve by substitution.

Huh what?

Paradoxica · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Huh what?

$$\noindent$ \begin{cases} y' + 2y = 10e^{2x} \\ y'' + 2y' = 20e^{2x} \end{cases}$

Eliminate the right hand side and then solve normally.

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
$$\noindent$ \begin{cases} y' + 2y = 10e^{2x} \\ y'' + 2y' = 20e^{2x} \end{cases}$

Eliminate the right hand side and then solve normally.

Lol that works (I'd hope) but that's so pointless, which was what I was calling the textbook

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Huh what?

I think he meant something like this?

$\noindent Given: $y' + 2y = 10e^{2x}$.$

$\noindent Differentiate: $y'' + 2y' = 20e^{2x}$.$

$\noindent Subtract twice the original ODE from the second: $y'' + 2y' -2y' -4y = 0 \Rightarrow y'' = 4y$. So $y= c_1 e^{2x}+ c_2 e^{-2x}$. Now, $y' = 2c_1 e^{2x} -2c_2 e^{-2x}$. Since $y'+2y = 10e^{2x}$, we have $\left(2+2\right)c_1 e^{2x} + 0e^{-2x} = 10e^{2x}\Rightarrow c_1 = \frac{5}{2}$. So the solution is $y = \frac{5}{2}e^{2x} + c e^{-2x}$ for some arbitrary constant $c$.$

Edit: outlined above.

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Lol that works (I'd hope) but that's so pointless, which was what I was calling the textbook

What was the context in which the Q. was asked? And did it say specifically to solve a characteristic equation etc., or did it simply say to solve the ODE? (So basically what did the Q. actually say?)

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
What was the context in which the Q. was asked? And did it say specifically to solve a characteristic equation etc., or did it simply say to solve the ODE? (So basically what did the Q. actually say?)

Yep.

Q43. Find the general solutions of
a)-d) all a bunch of random homogeneous second order ODEs

Q45. Solve:
a)-h) all a bunch of random non-homogeneous second order ODEs

All of which, are under a heading called "Section 3: Second order linear ordinary differential equations"

RenegadeMx · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Yep.

Q43. Find the general solutions of
a)-d) all a bunch of random homogeneous second order ODEs

Q45. Solve:
a)-h) all a bunch of random non-homogeneous second order ODEs

All of which, are under a heading called "Section 3: Second order linear ordinary differential equations"

homogenous is just using the char eqn
non-homogenous is same thing put u have to find that extra constant

nothing really too hard with these, some can be long unfortunately

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

I think he was just posting those to show the context.

Flop21 · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

How do you do the last unanswered question?

InteGrand · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How do you do the last unanswered question?

Rewrite your answer in terms of x. You got an answer of u, and x = 2*sinh(u), so get u in terms of x from this.

Flop21 · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Rewrite your answer in terms of x. You got an answer of u, and x = 2*sinh(u), so get u in terms of x from this.

ha yeah that's the question

I don't know how to get u = from that sinh(u) blah

InteGrand · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
ha yeah that's the question

I don't know how to get u = from that sinh(u) blah

x = 2*sinh(u). Therefore, sinh(u) = x/2. Then take the inverse sinh of both sides.

(It's basically the same procedure as if it had been x = 2*tan(u) or something instead – just need to utilise the inverse function.)

leehuan · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

So say I had to do this integral (under quite harsh timed conditions) and without a calculator.

$\int_0^2 \sqrt{x^2+4}dx$

Looking ahead, a trigonometric substitution would make me forced to integrate sec^3. So I naturally chose a hyperbolic substitution.

$$Let $x=2 \sinh t \implies dx = 2 \cosh t \,dt$

Using the trick of replacing cos with cosh in a trigonometric identity

$\begin{align*}\int_0^2 \sqrt{x^2+4}\,dx&=\int_0^{\sinh^{-1}2}\sqrt{4\cosh^2t}(2\cosh t dt)\\ &= 4\int_0^{\sinh^{-1}2}\cosh^2 t\, dt\\ &= 2\int_0^{\sinh^{-1}2}(1+\cosh 2t)dt\\ &= (2t+\sinh 2t) |_0^{\sinh^{-1}2}\end{align*}$

Obviously the substitution of 0 causes it to vanish. What's the best way to bust sinh(2 arcsinh 2)?

InteGrand · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
So say I had to do this integral (under quite harsh timed conditions) and without a calculator.

$\int_0^2 \sqrt{x^2+4}dx$

Looking ahead, a trigonometric substitution would make me forced to integrate sec^3. So I naturally chose a hyperbolic substitution.

$$Let $x=2 \sinh t \implies dx = 2 \cosh t \,dt$

Using the trick of replacing cos with cosh in a trigonometric identity

$\begin{align*}\int_0^2 \sqrt{x^2+4}\,dx&=\int_0^{\sinh^{-1}2}\sqrt{4\cosh^2t}(2\cosh t dt)\\ &= 4\int_0^{\sinh^{-1}2}\cosh^2 t\, dt\\ &= 2\int_0^{\sinh^{-1}2}(1+\cosh 2t)dt\\ &= (2t+\sinh 2t) |_0^{\sinh^{-1}2}\end{align*}$

Obviously the substitution of 0 causes it to vanish. What's the best way to bust sinh(2 arcsinh 2)?

Use double angle formula for sinh: sinh(2a) = 2sinh(a)cosh(a). Note cosh(arcsinh(u)) = sqrt(1 + sinh^2 (arcsinh(u)) = sqrt(1+u^2).

leehuan · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Use double angle formula for sinh: sinh(2a) = 2sinh(a)cosh(a). Note cosh(arcsinh(u)) = sqrt(1 + sinh^2 (arcsinh(u)) = sqrt(1+u^2).

Ahh I see

_____________

Algebra question next please

$\noindent\text{Prove that }S=\{\textbf{x}\in \mathbb{R}^3\mid 3x_1-4x_2=0\text{ and }2x_1+5x_3=0\}$ is a subspace of $\mathbb{R}^3$

For a question like this where there is multiple conditioning, what is perhaps the fastest way to do so? I was taught a method to parametrise the two linear equations and then prove the required three axioms (for the subspace theorem) from there.

$$i.e. I rewrote the equations as $\textbf{x}=\lambda \begin{pmatrix}-20\\-15\\8\end{pmatrix} \mid \lambda \in \mathbb R$

InteGrand · Aug 28, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Ahh I see
_____________

Algebra question next please

$\noindent\text{Prove that }S=\{\textbf{x}\in \mathbb{R}^3\mid 3x_1-4x_2=0\text{ and }2x_1+5x_3=0\}$ is a subspace of $\mathbb{R}^3$

For a question like this where there is multiple conditioning, what is perhaps the fastest way to do so? I was taught a method to parametrise the two linear equations and then prove the required three axioms (for the subspace theorem) from there.

$$i.e. I rewrote the equations as $\textbf{x}=\lambda \begin{pmatrix}-20\\-15\\8\end{pmatrix} \mid \lambda \in \mathbb R$

$\noindent Note $\mathbf{x}\equiv \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\in \mathbb{R}^3$ is in $S$ iff $A\mathbf{x} = \mathbf{0}_{\mathbb{R}^{2}}$, where $A$ is the matrix $\begin{pmatrix}3 & -4 & 0 \\ 2 & 0 & 5 \end{pmatrix}$ and $\mathbf{0}_{\mathbb{R}^{2}} = \begin{pmatrix} 0 \\ 0\end{pmatrix}$ is the zero-vector of $\mathbb{R}^{2}$. Hence $S$ is the \textsl{kernel} (aka \textsl{null space}) of the matrix $A$, so it is a subspace of $\mathbb{R}^{3}$.$

$\noindent (In general, for any given $m$-by-$n$ matrix $M$ over a field $\mathbb{F}$ (e.g. $\mathbb{R}$), the kernel of $M$, i.e. $S:=\left\{\mathbf{x}\in \mathbb{F}^{n} : M\mathbf{x}= \mathbf{0}_{\mathbb{F}^{m}} \right\}$, is a vector subspace of $\mathbb{F}^{m}$. To prove this, note that $S$ is a subset of a known vector space ($\mathbb{F}^{n}$) and the zero vector is in $S$. Moreover, $S$ is clearly closed under addition and scalar multiplication (prove this using matrix multiplication properties like distributivity etc.). Then by the `subspace lemma/theorem', $S$ is a subspace.)$

Flop21 · Aug 29, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Any old set of five vectors in $\mathbb{R}^{5}$ need not be a basis. E.g. if the set contains the zero vector, it won't be a basis (because it's automatically now linearly \emph{de}pendent).$

$\noindent The maximum number of linearly independent vectors we can have in a subset from $\mathbb{R}^{n}$ is $n$.[/B] So for instance, it's impossible to have a set of four vectors in $\mathbb{R}^{3}$ be linearly independent. If we have a set of \emph{more} than $n$ vectors from $\mathbb{R}^{n}$, this set is guaranteed linearly \emph{de}pendent. If we have $n$ or fewer vectors, we can't conclude whether they're linear independent just from this, we'd need to do further things (like row-reduce a matrix).$

Is this different for Pn ?

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

Kosovo is Serbian

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Kosovo is Serbian

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)