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Alpha_ProgDes

Double derivatives, taking it's integral

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this is my problem. y"-6y'+20y = 0 now i've gotten this far after i've taken the integral the first time ....-6y+10y^2 but i have no idea what to do with the y". if i remember correctly it is equivalent to d^2y. i don't think the integral of y" is ydy... is it? a little help please. and for the record no this is not my homework. i passed Calc II a long time and Diff Eq. last year. too bad i sold my book edit: gave complete equation. no other variables then just what's up there. [edited by - Alpha_ProgDes on September 3, 2003 12:03:06 AM]

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Guest Anonymous Poster
I''ll assume your equation equals zero.

Because this is an ordinary second-order equation with constant coefficients, we may use the characteristic polynomial (r^2 - 6r + 20). The roots for this poly can be found by using the Quadratic Formula.

The general solution is then just

A * e^(r1 * t) + B * e^{r2 * t)

Where r1 and r2 are the roots to the characteristic polynomial and A and B are constants whose values depend on the initial conditions of your system. The variable t, of course, is just the independent variable (you didn''t specify if it was x or t).

Hope that helps!

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Exactly as AP says. Google ( or use a text book ) for the method of undetermined coefficients for more details.

Edit: That would be variation of parameters even, damn my memory for names of methods.

[edited by - Keem on September 4, 2003 4:54:35 AM]

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yeah after Googling for about 30 minutes I finally found a worthwhile site. Damn brain... it''s need a defrag

anyway i saw something interesting on one page.
and i''m wondering why this wouldn''t work.
it looks sound...

y"-6y''+20y = 0

y" = 6y''-20y

take the integral (1st time)
y'' = 6y - 10y^2

take the integral (second time)
y = 3y^2 - (10/3)y^3


y(-(10/3)y^2 + 3y - 1) = 0


...hmmm, self-realization kicking in...
(taking root by Quad. Eq. gives real and imaginary roots...)
(will solve rest after class)

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Guest Anonymous Poster
quote:
Original post by Alpha_ProgDes
y" = 6y''-20y

take the integral (1st time)
y'' = 6y - 10y^2


You need to realise what variable you are integrating. It is not y, thus y²/2 is no primitive function to y (unless y(t) = t).

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homogeneous ( spleeing ) second order integral equation,
after solving that, passage are a little involved , anyway this euqation gives complex and coniugate solution , i try to explain

y''''-6y''+20y=0

now solve the for ''general'' quantity this gives

y^2-6y+20 =0

this euqation gives two values

y1= -3+ i sqrt( 11 )
y2= -3 -i sqrt ( 11 )

let alpha = -b/2 and betha = sqrt ( -delta ) / 2

the solution is

y = c1 cos(betha)*e^(alpha*x)+c2*sin(betha)*e^(-alpha*x)

this is a energy - transition state , toward nominal conditions

you have to solve Cahcy''s problem to get the
values for c1 and c2
normally this is done , with initial conditions.

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quote:
Original post by v71
homogeneous ( spleeing ) second order integral equation,
after solving that, passage are a little involved , anyway this euqation gives complex and coniugate solution , i try to explain

y''-6y''+20y=0

now solve the for ''general'' quantity this gives

y^2-6y+20 =0

this euqation gives two values

y1= -3+ i sqrt( 11 )
y2= -3 -i sqrt ( 11 )

let alpha = -b/2 and betha = sqrt ( -delta ) / 2

the solution is

y = c1 cos(betha)*e^(alpha*x)+c2*sin(betha)*e^(-alpha*x)

this is a energy - transition state , toward nominal conditions

you have to solve Cahcy''s problem to get the
values for c1 and c2
normally this is done , with initial conditions.



i did the general solution: y^2 + 6y + 20
and using the quadratic equation i got:
(Q.E. being: (-b +/- sqrt(b^2 - 4ac))/2a)

[6 +/- 2i(sqrt(11))] / 12
which simplifies to
[3 +/- i(sqrt(11)] / 6
and my general equation (the solution) is:

y = e^(-.5x)*[A cos (sqrt(11)x/6) + B cos (sqrt (11)x/6)]

did i do my math wrong?
i saw no mistakes.

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your math is not wrong , but there are 3 form to express a complex number, algebric , polar , exponential form
just choose the one you like most i use exponential because
normally i use to do laplace space algebric computation and the i revert back them into time domain if a oscillator function is needed to be studied in function of time,
P.S. maybe i did some mistake in the computation of factors, sorry ...

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