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anonymous
 5 years ago
how would i solve (x double dot) + x(x dot)+x=0
anonymous
 5 years ago
how would i solve (x double dot) + x(x dot)+x=0

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you can solve it by reduction of order

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you deleted that? it wasnt the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that one can be solved differently

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought reduction of order was for coefficients which are random polynomials of in t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x \prime \prime+xx \prime+x=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for this you say let u=dx/dt than d^2x/dt^2=u du/dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0than the equation will be u du/dt + xu+x=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh, that is clever, i will try that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and I guess you can use an integrating factor to solve this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now I have to go to the shop. If you have a problem just post it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sweet, thanks a ton you are amazing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that integral doesn't make sense, it is with respect to the wrong variable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have to change u back to x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did not solve it, what did you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah, i'm saying even setting up the integration factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dot means that we differentiate with respect to t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x is a function of t, but the integral should be with respect to t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea, i set up \[u \prime+ux+x=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u du/dt + xu+x=0 du/dt +u/x=x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u'=u du/dt by the chain rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0e^integral1/x is the integrating factor that is just x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so multiplying with x gives xdu/dt+u=x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but when you set up that equation du/dt +u/x=x we should be integrating with respect to t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are integrating with respect to x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the integrating factor no

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is only for making it a full derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the integrating factor makes sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it is like this, but I would not bet on it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the actual integratioin though does not and why is the equation u du/dt rather than du/dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u=dx/dt u dot= x double dot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this part is not 100% clear for me :) but I am sure that is right and comes from the chain rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just posted a question about it, maybe some1 is wise

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(I'm just a 1st year maths student)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, thank you for your help, i will check back later. i really appreciate it andras
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