Waching daily Sep 3 2018

Hey guys, and welcome back to the School Finesse podcast. I'm your host, Amen.

Today, we have tip #3. Here's the tip.

Understand that your teacher, your professor, your instructor is available to help you.

You know, so many times we

students write-off the teacher and assume that

they are too busy to help. They won't understand. They don't even care. But

these are all terrible assumptions.

Here's a sub tip: establish communication with your professor.

If you don't understand the assignment, notify the professor.

If you're going to be late for class, notify the professor.

If you need assistance for making a decision, notify the professor.

That professor may be able to direct you in the right path or to someone who can help you.

Do. not. be. afraid. to. talk. to. the. teacher.

The teacher is there to help.

But also remember, the teacher may be more willing to help you if you are helping yourself.

So what do I mean by you have to help yourself?

I mean you have to take initiative and email your professor, contact them, let

them know the issues you are having so they can help you in the best way possible.

Above all, remember you have to put your best foot forward and remember

that your professor, your instructor, your teacher is a resource to help you succeed.

CHALLENGE FOR THE WEEK

Can you share 1 positive experience that

you've had with a professor or an instructor in the past that was beneficial to you?

Share your thoughts with us in the comment section.

Your comment may encourage someone else.

Now, on to the News Fix.

So last week, students were still starting their first day back to school.

And you know, on the first day of school, you may have some concerns.

Who are your new teachers? How will you get from one class to the other?

But maybe more importantly, for some, who are you going to sit with during lunch time?

This is a big deal regardless if you are in elementary, middle, or high school, or maybe even college.

Well, at Boiling Springs High School in South Carolina, there was a student named Andrew.

He was a junior and normally sits alone at lunch without anyone approaching him to join their table.

Well his first day of school last week was different.

Students from the school council approached him and a couple of other students who were alone, and they invited them to their table.

They ended up having a conversation, talking and everything went well.

And they ended up going to the movies during the end of the week.

Now Andrew finally feels like he belongs and he wants to go to school.

And his mother says that this act of kindness toward her son was an answered prayer.

Also last week, a kindergartner slipped out of his classroom at about 12:20 p.m and decided to head home for the day.

Now, I don't remember too much about kindergarten, but I think I'd be a little

scared to figure out my own way home. Kudos to the smart kid.

Of course, his parents are - aren't happy that their child was able to slip out of the school so easily.

So the school system has apologized and has vowed to tighten up their accountability and safety practices.

That's all for this week's News Fix. Links to the news articles will be in a description.

Be sure to subscribe, like, comment, and share with a friend.

Always remember, tune in every Monday morning for a new tip on how you can finesse your way through school this week.

I'm your host, Amen, signing off.

For more infomation >> Your teacher is available! School Finesse Podcast - Ep. 3 - Duration: 3:51.

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For more infomation >> Robocar Poli | There Is A Ghost In The Cave | Toys For Kids - Duration: 4:08.

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2nd Order Linear Differential Equations P I is Trigonometry - Duration: 10:21.

Hi, welcome to another tutorial

in my series on

second order

Linear differential equations

which equals some function of x.

Now in the first tutorial in this series

I showed you that the general solution was

over for y equals complementary function plus particular integral

C.F. + P.I. for short

and in this video I'm going to be looking

at equations where f(x) is a trigonometric function

and in this example we could find the general solution to the equation

d²y/dx² + 2dy/dx - 3y = 6cos2x

we are given that the complementary function for this equation

is Ae^-3x

plus Be^x, if you are unsure

how we get this complementary function just click on this link up here

and it will go back to our tutorial that

showed you how we got this okay, so

first of all we need to work out what the particular integral is going to be, so

P.I. then for short. Now we'll be dealing with

trigonometric functions of the form say

mcosnx, if you like

in this example 6cos2x then

whether it's 6cos2x even 6sin2x

then the particular integral that we use is to let y

equal any constant say λ

then we have cos2x but it doesn't stop there

we put plus another constants let's say μ

and then we multiply that with

sin2x. So I could have had say

6sin3x here or cos2x

6sin3x my particular integral

would have be in the form

y = λcos3x + μsin3x

so as I said this is another way you got cosine here or sine

go to this style.

Okay, well on that basis we

do as we done with some many times before in these types of questions if you've been

looking at them and that is to find dy/dx

and then the second differential and we substitute those

values into the equation. Okay so for this one if we differentiate

y with respect to x for this first time here

we're going to get

-2λsin2x and then when we differentiate

the second term here with respect to x we're going to get

plus 2μcos2x

Okay, since dy/dx now we need the second differential

d²y/dx²

so differentiating the first time here

is going to give us

-4λcos2x and then if we differentiate this term here

we're going to get

-4μsin2x. Okay so

we've got ydy/dx and d²y/dx².

Now because there's lack of room here

what I'm going to do is will substitute

these values into this equation up here will call it one

okay, so sub into one

but what we're going to have now?

is that for this term

-3y is just triple

all these values here so we've got that -3y

equals

times this by -3 we are going to get

-3λcos2x and if we times

this term by -3 we get

-3μsin2x.

Next, we've got to

add 2logs of dy/dx, so

if we do 2logs of dy/dx

2dy/dx or we are going to get for this form

well if we double each of these terms what I am going to do

is take this term first so that I can working columns here

and will have this first column as the cos2x

and terms, so what double this term

so we've got 4μcos2x

4μcos2x

and if I double this term we're going to have

-4λsin2x

and finally we've got d²y/dx². We don't have to do anything to

that

so let's just copy that in

d²y/dx² we'll take it in the order

it's there so we got

-4λcos2x and then

-4μsin2x.

Okay, well if we now

add these terms together okay we add these terms together

then we can see that if we compare the cos2x terms

looking at this first column here we've got a total of

-7λ + 4μ

and that must be equal to 6.

So, just border this off. Okay,

and will say we'll compare

because 2x terms, we look at the coefficient there

so as say we got -3λ, -4λ & the last -7λ

and then plus the 4μ

and that must be equal to 6 here

this also compare now the coefficients of

sin2x so just write that here compare

sin2x coefficients and

again if we do this what we've got

is -4λ

and for the μ we've got -3μ, -4μ that's -7μ

and we've got no terms on the right hand side

that's sin2x the coefficients the values would be

0 then. Now if you were to solve these two simultaneous equations for λ

μ you'd find that λ turns out to be

-42/65

I'll leave it up to you to just work that out

and μ will turn out to be

24/65

so what that means now is

that the particular integral

P.I. for sure then is going to be

or we call it is λcos2x, so it's going to be

-42/65cos2x

and then

we got plus μsin2x that is plus

24/65 of sin2x

and that gives us a particular integral.

So what we have next is the

general solution or just abbreviate that's G.S.

and that's going to be the complementary function which is

Ae to the power -3x plus

Be to the power x and into that we just have to

add a particular integral.

So instead of being plus we've now already got minus

42/65 then

of cos2x and then

plus 24/65

of sin2x

okay so how this give you an idea now

how we can handle trig-types

of the form where you've got a constant say

multiply by cos of some multiple angle

or this would work just as equally while

if it was a constant times Sin of some multiple angle

still take this kind of form for particular integral.

Well it brings us now to the end of this tutorial.

For more infomation >> 2nd Order Linear Differential Equations P I is Trigonometry - Duration: 10:21.

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Lexus IS 300H F SPORT LINE Schuifdak Navi Keyless Go camera F1 - Duration: 1:10.

For more infomation >> Lexus IS 300H F SPORT LINE Schuifdak Navi Keyless Go camera F1 - Duration: 1:10.

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2nd Order Linear Differential Equations P I is Constant - Duration: 8:19.

Hi, now if you would be

watching my earlier videos in the series.

We were looking at

solving a second order linear

differential equation something like this but it equals 0.

And the second order linear differential equation

was where these values here a, b and c

were constants, now in this video what I want to do

is introduce you to equations like this

where they are equal to not 0 but some function of x,

and it can be shown that

the general solution of these types of equations

essentially are made up of what we call a complementary function plus a Particular Integral.

So, what do we mean by complementary function?

When a complimentary functional of C.F. for short

is a solution to this particular

type of second order linear differential equation,

you can see it equals 0 and we handled these as said

in earlier tutorials, if you're not sure how to handle these

just click on this link here and it will take you back to that series.

So, we need to

work out the solution to this type of equation that be a complementary

function,

and to this we have to add

what we call a particular integral, so what do we mean by

a particular integral? Well, the particular integral

or P.I. for short, is a solution

to this particular equation a value that works. Okay.

Now, this way I can demonstrate this is through an example

this going to be a fairly straight forward example

first of all just to get you into the swing of this

and we'll tackle harder examples later on,

so the example going to look at then

is this one, where we got to solve

d<SUP>2</SUP>y/dx<SUP>2</SUP> + 2dy/dx - 3y = 4,

and

we can see that it's got this particular

format a function of x is

a constant. So first of all, what we've got to

workout is the complementary function.

The complimentary function or C.F. for short which is put this down here C.F.

is just highlight that okay, the complementary function then

is where we solve the equation

equalling 0, so we take the left hand side of this and so we've got

d<SUP>2</SUP>y/dx<SUP>2</SUP> + (2dy/dx) - 3y = 0.

Now, when we solve this earlier, we had to work out what the

Auxiliary equation was okay, so

which is put that down here the Auxiliary Equation

and that Auxiliary Equation

would be 1m<SUP>2</SUP> or just the m<SUP>2</SUP> + 2m - 3 = 0,

so we've got

m<SUP>2</SUP> + 2m - 3 = 0

and we would solve this quadratic equation and in this instance

this one

factorizes, the factorize is to (m + 3)

multiply by (m - 1) and it equals 0

leading to two roots, two real and different roots

and m = -3 and

m = 1 and I showed you

in the video for solving these

that if we have real in different roots the general solution

for that was something of the form y equals

a constant that's called

Ae^-3x + and then another constant

B say, e^1x which is simply

e<SUP>x</SUP> okay, so this would be

our complementary function this is x up there

okay, let's just move on what we've got to do now

is calculate the particular integral

P.I. we will call it, they gave for short.

So for the particular integral

this is the bit that will change when you're doing these kind of questions

we are going into more depth on this in later tutorials

but for now what we do is we take

a trial value, a trial value for y

and it's going to vary depending on the question

when you are handling a constant like this

we just say let y equal some value let's say λ

λ being a constant, what we do,

is we now find out what dy/dx would be

well dy/dx would be equal

0 for this value, we differentiated the constant

with respect to x and if we

differentiate it again d<SUP>2</SUP>y/dx<SUP>2</SUP>

well that too

is going to be 0, so

if we substitute these values, these three values

into this equation here let's call it one and say

sub into one okay,

let's see what we got. So, d<SUP>2</SUP>y/dx<SUP>2</SUP> is equal to 0, so we just put that

down here 0,

then we got plus 2 times dy/dx that's two times zero again

I know 0 but I'm just purposely putting that in there to show that

you can refer back to what I've been doing and then -3 times y

3 times y, y is λ

and we told it equals 4

so that means that -3λ = 4

so therefore λ must be equal

-4/3

so what that means is that therefore our solution

our general solution to our equation here

is the complementary function plus the particular integral.

So what we've got then is therefore y

equals the complementary function

Ae<SUP>-3x</SUP> + Be<SUP>x</SUP>

and then plus our particular integral, so adding -4/3 says just simply

-4/3

So there you go fairly straightforward I hope

but the difficult if anything is going to come about

when we look at this particular integral

because as I said earlier it's going to vary depending on the more function of x

you've got here,

we're going to look at linear functions,

we're going to look at exponential functions

and trig functions in this series.