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

- anonymous

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

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- anonymous

help

- anonymous

I think you can solve it by reduction of order

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## More answers

- anonymous

how?

- anonymous

you deleted that? it wasnt the same

- anonymous

yea, i had it wrong

- anonymous

that one can be solved differently

- anonymous

i thought reduction of order was for coefficients which are random polynomials of in t

- anonymous

\[x \prime \prime+xx \prime+x=0\]

- anonymous

for this you say let u=dx/dt
than
d^2x/dt^2=u du/dt

- anonymous

than the equation will be
u du/dt + xu+x=0

- anonymous

ahh, that is clever, i will try that

- anonymous

thanks a ton

- anonymous

and I guess you can use an integrating factor to solve this

- anonymous

now I have to go to the shop. If you have a problem just post it

- anonymous

sweet, thanks a ton you are amazing

- anonymous

that integral doesn't make sense, it is with respect to the wrong variable

- anonymous

you have to change u back to x

- anonymous

I did not solve it, what did you get?

- anonymous

ah, i'm saying even setting up the integration factor

- anonymous

dot means that we differentiate with respect to t

- anonymous

x is a function of t, but the integral should be with respect to t

- anonymous

yea, i set up \[u \prime+ux+x=0\]

- anonymous

u du/dt + xu+x=0
du/dt +u/x=-x

- anonymous

u'=u du/dt by the chain rule

- anonymous

e^integral1/x is the integrating factor
that is just x

- anonymous

so multiplying with x gives
xdu/dt+u=-x^2

- anonymous

(xu)dot=-x^2

- anonymous

xu=-(x^3)/3 +C

- anonymous

u=-(x^2)/3 +c/x

- anonymous

u=dx/dt

- anonymous

dx/dt=-(x^2)/3 +c/x

- anonymous

so t= -(x^3)/9 +Clnx

- anonymous

but when you set up that equation du/dt +u/x=-x we should be integrating with respect to t

- anonymous

you are integrating with respect to x

- anonymous

for the integrating factor no

- anonymous

it is only for making it a full derivative

- anonymous

the integrating factor makes sense

- anonymous

I think it is like this, but I would not bet on it :-)

- anonymous

the actual integratioin though does not and why is the equation u du/dt rather than du/dt

- anonymous

u=dx/dt
u dot= x double dot

- anonymous

this part is not 100% clear for me :) but I am sure that is right and comes from the chain rule

- anonymous

I just posted a question about it, maybe some1 is wise

- anonymous

(I'm just a 1st year maths student)

- anonymous

well, thank you for your help, i will check back later. i really appreciate it andras

- anonymous

your welcome

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