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

- anonymous

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- anonymous

##### 1 Attachment

- Adi3

What do you think??

- Adi3

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## More answers

- anonymous

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

- Adi3

@IrishBoy123 is that right??

- anonymous

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

- Adi3

bump your question

- Adi3

- anonymous

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.

- anonymous

I wonder if this has something to do with Laplace

- IrishBoy123

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 \(y = A e^x + B e^{2x}\)

- IrishBoy123

@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

- anonymous

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 ?

- Adi3

can some one medal me pls.

- IrishBoy123

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

- anonymous

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

- IrishBoy123

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

- IrishBoy123

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

- IrishBoy123

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.

- Adi3

thank you for giving me a medal.

- Adi3

appreciate it.

- Adi3

:(

- Adi3

Thank you, i have 49 C now

- Adi3

1 more medal for 50.

- Adi3

1 more medal for 50.

- anonymous

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

- anonymous

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

- IrishBoy123

yes!

- anonymous

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

- anonymous

n_1 = 1, n_2 = 2

- anonymous

so Ae^x + Be^2x

- anonymous

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

- IrishBoy123

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

- anonymous

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

- IrishBoy123

yes you are seeing the pattern emerge.

- anonymous

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?

- anonymous

Like this;

##### 1 Attachment

- anonymous

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

- IrishBoy123

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

- IrishBoy123

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

- anonymous

Ah, much easier! Yes - I got to Ae^wit + Be^-wit
How do we proceed ?

- IrishBoy123

\(\theta = A e^{i \omega \; t} + B e^{-i \omega\; t}\)
\(= A ( \cos \omega t + i \sin \omega t) + B ( \cos (-\omega t) + i \sin (-\omega t)) \)
\(= A ( \cos \omega t + i \sin \omega t) + B ( \cos \omega t - i \sin \omega t) \)
\( = (A + B) \cos \omega t + i( A - B ) \sin \omega t \)
\(= C \cos \sqrt{\dfrac{g}{R}} t + D \sin \sqrt{\dfrac{g}{R}} t \) where \(C, D = const.\)

- anonymous

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

- IrishBoy123

eulers formule
\[e^{ix}=\cos x+i\sin x\]
https://en.wikipedia.org/wiki/Euler%27s_formula

- anonymous

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

- anonymous

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?

- anonymous

Apparently the answer is supposed to be

##### 1 Attachment

- IrishBoy123

no
here \(\alpha = 0\) so you have \(y =1 \cdot (A \sin \beta x + B \cos \beta x)\)
and \(\beta = \omega = \sqrt{\dfrac{g}{R}}\)

- IrishBoy123

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.

- IrishBoy123

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

- anonymous

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