Question

2. degree diff. equation?
I have honestly forgot more or less everything about 2. degree diff. equations.

How do I solve this equation? (posted below)
Note that I am not after "the answer". I am after the steps taken towards getting an answer - so that I may (hopefully) recall how to do it

Answer 1

What do you think??

Answer 2

@Ivanskodje

Answer 3

I think:
Kx^2 + 9.81 - which has no real roots so it will be imaginary

Then I think I need to find another something and put it into an ... "general"-somethingsomething

Answer 4

@IrishBoy123 is that right??

Answer 5

I meant x^2 + 9.81 (or PHI or whatever symbol)

Answer 6

bump your question

Answer 7

@IrishBoy123

Answer 8

x1 = root(9.81) * i
x2 = -root(9.81) * i

Although I have no idea what to do with this now. I think there was a formula or something somewhere.

Answer 9

I wonder if this has something to do with Laplace

Answer 10

let's take a different example as @Ivan wants a general approach and not the answer

say \(y'' - 3y' + 2y = 0\)

if we use the differential operator \(D = \dfrac{d}{dx}\) to simplify the algebra we have

\((D^2 - 3 D + 2)y = 0\)
\((D-2)(D-1)y = 0\)

taking the first root: \((D-1)y = 0 \implies y'-y = 0\)
so \(y' = y\)
\(\dfrac{dy}{y} = dx \)
\(\ln y = x + c\)
\(y = e^{x+c} = Ae^x\)



by the same logic
\((D-2)y = 0 \implies y'-2y = 0\)
and gives \(y = Be^{2x}\)

to the solution is \(y = A e^x + B e^{2x}\)

Answer 11

@Ivanskodje , if you follow that, we have a way to solve the one you have posted. just pattern match, some complex exponentials will crop up but we can deal with that...

of course we can also use laplace transform if you like. normally you'd save laplace for initial value problems...but we can do it if you are familiar with laplace

Answer 12

D = d/dt - However, isn't y'' d^2y / dt^2 ? I mean... or ... \[y ^{''} = \frac{ d^2y }{ dt^2 } -> D=\frac{ d }{ dt } \]
We set D to be d/dt, how can y'' be D^2 if the d below isn't ^2 as well?

Or is the dt^2 = (dt)^2 ?

Answer 13

can some one medal me pls.

Answer 14

\(D^2 \implies \dfrac{d}{dx}\left( \dfrac{d}{dx} \right) = \dfrac{d^2}{dx^2}\)

it's an algebraic shortcut

the alternative is to assume straight out that the solutions to the equation that i gave as an example come in the form \(y = A x^{\ x} + B x^{2 x}\) ie having found the roots of the underlying equation.

i personally think that less gratifying but that is called the method of undetermined coefficients

you can find loads of it here:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 15

Although shou... oooh! "dx" below is like one unit? Not d*x right?

Answer 16

and if you're interested:
https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/undetermined-coefficients/v/undetermined-coefficients-1

Answer 17

\[D^2 (y) = \dfrac{d^2 y}{dx^2}\]

Answer 18

or
\(D^2 (\theta) = \dfrac{d^2 \theta}{dt^2}\), if \(\theta\) and \(t\) are the dependent and independent variable, as seems to be the case in your DE.

Answer 19

thank you for giving me a medal.

Answer 20

appreciate it.

Answer 21

:(

Answer 22

Thank you, i have 49 C now

Answer 23

1 more medal for 50.

Answer 24

1 more medal for 50.

Answer 25

I am trying to understand the e^(x+c) = Ae^x - How do we figure it is equal to Ae^x? Formula?

Answer 26

Because e^c is a constant - so A? I think I got it

Answer 27

yes!

Answer 28

On the second part; (D−2)y=0 - Where did the B come from? I got to e^2t. Do we just assume it is multiplied with some unknown constant?

There is a formular that is like y = Ae^(n_1)x + Be^(n_2)x .... hmm

Answer 29

n_1 = 1, n_2 = 2

Answer 30

so Ae^x + Be^2x

Answer 31

Applies when n_1 is not equal to n_2, but are real

Answer 32

\( (D−2)y=0 \implies y' = 2y\)
\(\dfrac{dy}{y} = 2 dx\)
\(\ln y = 2x + c\)
\(y = e^{2x+c} = Be^{2x} \)

Answer 33

oh of course! Forgot the C... never forget the C!

Answer 34

yes you are seeing the pattern emerge.

Answer 35

Now, when going back to the task at hand; I see that because we get an unreal number1 and number2 - there is a different way to progress? According to a book; the two numbers I get, somehow becomes n_1 = a+ib and n_2 = a-ib.

However, when I solve for O, I get root(9.81) * i and -root(9.81)*i. How do that translate?

Answer 36

Like this;

Answer 37

The one to the right is the "format" I need in order to use the second formula where I use sin and cos

Answer 38

well, leave the actual numbers out of it until we solve.

we have \(\ddot \theta + \omega^2 \theta = 0\) where \(\omega^2 = \dfrac{g}{R}\)

[that \(\omega^2 \) substitution (??),....., just go with it and you'll see why it's handy]

\((D^2+\omega^2)\theta = 0\)
\(D = \pm i \omega\)

thus as per above pattern, we have
\(\theta = A e^{i \omega \; t} + B e^{-i \omega\; t}\)

so we have the complex solution you mention but the real part is zero. we are left with \(i \omega\) in the solution

Answer 39

we then switch that to sin and cos, when you are ready....

Answer 40

Ah, much easier! Yes - I got to Ae^wit + Be^-wit

How do we proceed ?

Answer 41

\(\theta = A e^{i \omega \; t} + B e^{-i \omega\; t}\)

\(= A ( \cos \omega t + i \sin \omega t) + B ( \cos (-\omega t) + i \sin (-\omega t)) \)

\(= A ( \cos \omega t + i \sin \omega t) + B ( \cos \omega t - i \sin \omega t) \)

\( = (A + B) \cos \omega t + i( A - B ) \sin \omega t \)

\(= C \cos \sqrt{\dfrac{g}{R}} t + D \sin \sqrt{\dfrac{g}{R}} t \) where \(C, D = const.\)

Answer 42

That looks very familiar - I just need to find a reference to the convertion between e^(...) to cos(...)+isin(...)

Working on it now

Answer 43

eulers formule

\[e^{ix}=\cos x+i\sin x\]

https://en.wikipedia.org/wiki/Euler%27s_formula

Answer 44

I just need to find a reference to it in the formula book - and put the pieces together. I find something similar but not quite the same - working on it

Answer 45

This is the formula I have to work with;

\[\lambda_1 = \alpha +i \beta, \lambda_2 = \alpha -i \beta\]

\[y = e ^{\alpha x} (A \sin \beta x + B \cos \beta x)\]

I assume this makes the alpha 0 and beta w ?
Or would this be the wrong direction?

Answer 46

Apparently the answer is supposed to be

Answer 47

no

here \(\alpha = 0\) so you have \(y =1 \cdot (A \sin \beta x + B \cos \beta x)\)

and \(\beta = \omega = \sqrt{\dfrac{g}{R}}\)

Answer 48

yes, you also can use a angle formula to express it as a cosine or a sine with a phase shift \(\phi\).

so \(\sin (\omega t + \phi) = \sin \omega t \cos \phi + \cos \omega t \sin \phi\)

again you can pattern match back into the solution above

these are all ways of expressing the solution, you take your pick.

Answer 49

so\[A \sin (\omega t + \phi) = A [\sin \omega t \cos \phi + \cos \omega t \sin \phi] \\= B \sin \omega t + C \cos \omega t \]

Answer 50

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