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help

I think you can solve it by reduction of order

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