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anonymous
 one year ago
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
 one year ago
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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 is that right??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I meant x^2 + 9.81 (or PHI or whatever symbol)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x1 = 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
 one year ago
Best ResponseYou've already chosen the best response.0I wonder if this has something to do with Laplace

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3let'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 \((D^2  3 D + 2)y = 0\) \((D2)(D1)y = 0\) taking the first root: \((D1)y = 0 \implies y'y = 0\) so \(y' = y\) \(\dfrac{dy}{y} = dx \) \(\ln y = x + c\) \(y = e^{x+c} = Ae^x\) by the same logic \((D2)y = 0 \implies y'2y = 0\) and gives \(y = Be^{2x}\) to the solution is \(y = A e^x + B e^{2x}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3@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
 one year ago
Best ResponseYou've already chosen the best response.0D = 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
 one year ago
Best ResponseYou've already chosen the best response.1can some one medal me pls.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\(D^2 \implies \dfrac{d}{dx}\left( \dfrac{d}{dx} \right) = \dfrac{d^2}{dx^2}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0Although shou... oooh! "dx" below is like one unit? Not d*x right?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3and if you're interested: https://www.khanacademy.org/math/differentialequations/secondorderdifferentialequations/undeterminedcoefficients/v/undeterminedcoefficients1

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\[D^2 (y) = \dfrac{d^2 y}{dx^2}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3or \(D^2 (\theta) = \dfrac{d^2 \theta}{dt^2}\), if \(\theta\) and \(t\) are the dependent and independent variable, as seems to be the case in your DE.

Adi3
 one year ago
Best ResponseYou've already chosen the best response.1thank you for giving me a medal.

Adi3
 one year ago
Best ResponseYou've already chosen the best response.1Thank you, i have 49 C now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to understand the e^(x+c) = Ae^x  How do we figure it is equal to Ae^x? Formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because e^c is a constant  so A? I think I got it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On 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
 one year ago
Best ResponseYou've already chosen the best response.0Applies when n_1 is not equal to n_2, but are real

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\( (D−2)y=0 \implies y' = 2y\) \(\dfrac{dy}{y} = 2 dx\) \(\ln y = 2x + c\) \(y = e^{2x+c} = Be^{2x} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh of course! Forgot the C... never forget the C!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yes you are seeing the pattern emerge.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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 = aib. However, when I solve for O, I get root(9.81) * i and root(9.81)*i. How do that translate?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The one to the right is the "format" I need in order to use the second formula where I use sin and cos

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3well, 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
 one year ago
Best ResponseYou've already chosen the best response.3we then switch that to sin and cos, when you are ready....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, much easier! Yes  I got to Ae^wit + Be^wit How do we proceed ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\(\theta = A e^{i \omega \; t} + B e^{i \omega\; t}\) \(= A ( \cos \omega t + i \sin \omega t) + B ( \cos (\omega t) + i \sin (\omega t)) \) \(= A ( \cos \omega t + i \sin \omega t) + B ( \cos \omega t  i \sin \omega t) \) \( = (A + B) \cos \omega t + i( A  B ) \sin \omega t \) \(= C \cos \sqrt{\dfrac{g}{R}} t + D \sin \sqrt{\dfrac{g}{R}} t \) where \(C, D = const.\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That looks very familiar  I just need to find a reference to the convertion between e^(...) to cos(...)+isin(...) Working on it now

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3eulers formule \[e^{ix}=\cos x+i\sin x\] https://en.wikipedia.org/wiki/Euler%27s_formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0Apparently the answer is supposed to be

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3no 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
 one year ago
Best ResponseYou've already chosen the best response.3yes, 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
 one year ago
Best ResponseYou've already chosen the best response.3so\[A \sin (\omega t + \phi) = A [\sin \omega t \cos \phi + \cos \omega t \sin \phi] \\= B \sin \omega t + C \cos \omega t \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Processing information... ! Still working on it
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