how would i solve (x double dot) + x(x dot)+x=0

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how would i solve (x double dot) + x(x dot)+x=0

Mathematics
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I think you can solve it by reduction of order

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Other answers:

how?
you deleted that? it wasnt the same
yea, i had it wrong
that one can be solved differently
i thought reduction of order was for coefficients which are random polynomials of in t
\[x \prime \prime+xx \prime+x=0\]
for this you say let u=dx/dt than d^2x/dt^2=u du/dt
than the equation will be u du/dt + xu+x=0
ahh, that is clever, i will try that
thanks a ton
and I guess you can use an integrating factor to solve this
now I have to go to the shop. If you have a problem just post it
sweet, thanks a ton you are amazing
that integral doesn't make sense, it is with respect to the wrong variable
you have to change u back to x
I did not solve it, what did you get?
ah, i'm saying even setting up the integration factor
dot means that we differentiate with respect to t
x is a function of t, but the integral should be with respect to t
yea, i set up \[u \prime+ux+x=0\]
u du/dt + xu+x=0 du/dt +u/x=-x
u'=u du/dt by the chain rule
e^integral1/x is the integrating factor that is just x
so multiplying with x gives xdu/dt+u=-x^2
(xu)dot=-x^2
xu=-(x^3)/3 +C
u=-(x^2)/3 +c/x
u=dx/dt
dx/dt=-(x^2)/3 +c/x
so t= -(x^3)/9 +Clnx
but when you set up that equation du/dt +u/x=-x we should be integrating with respect to t
you are integrating with respect to x
for the integrating factor no
it is only for making it a full derivative
the integrating factor makes sense
I think it is like this, but I would not bet on it :-)
the actual integratioin though does not and why is the equation u du/dt rather than du/dt
u=dx/dt u dot= x double dot
this part is not 100% clear for me :) but I am sure that is right and comes from the chain rule
I just posted a question about it, maybe some1 is wise
(I'm just a 1st year maths student)
well, thank you for your help, i will check back later. i really appreciate it andras
your welcome

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