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OpenStudy (anonymous):

How to find the general solution of a second order linear equation given three particular solutions? Ly=f => This means it's a non-homogeneous right? y=x , y=e^x+x , y=2e^x+x+1 are the three given particular solutions.

OpenStudy (experimentx):

add them up ... and also add the solution of the Ly=0 (homogeneous part)

OpenStudy (anonymous):

how do you find the the solution of the homogeneous part?

OpenStudy (anonymous):

y=yp+th => that's the form I would follow right?

OpenStudy (experimentx):

you need to know what L is ...

OpenStudy (anonymous):

*yp+yh

OpenStudy (anonymous):

it's not given though...

OpenStudy (anonymous):

the problem just say Ly=f

OpenStudy (experimentx):

yeah ... sometimes it is also called complimentary function.

OpenStudy (anonymous):

Here's the complete problem: Three particular solutions of a certain second order linear equation Ly=f are y=x, y=e^x+x, y=2e^x+1+x. What is the general solution?

OpenStudy (anonymous):

This problem from the section talking about the Wronskian

OpenStudy (experimentx):

these particular solutions are due to superposition y=x , y=e^x+x , y=2e^x+x+1 probably your question just expects to write \[ c_f + y_p\] where cf is complimentary function.

OpenStudy (anonymous):

I don't exactly know what complimentary function means... Could you explain it to me/?

OpenStudy (anonymous):

so my yp would = 3e^x+3x+1?

OpenStudy (experimentx):

cf is just solution to homogeneous part.

OpenStudy (anonymous):

the solution to Ly=0, correct? And how do I find that...?

OpenStudy (experimentx):

I don't know how wronskian is related to this ... probably with variation of parameters. though if there.

OpenStudy (anonymous):

so should my answer just be y=cf+3e^x+3x+1?

OpenStudy (experimentx):

without L given , I don't think i can find out. Let me see if ... i can extract info out of variation of parameters.

OpenStudy (anonymous):

ok, thank you.

OpenStudy (experimentx):

man .. this is crazy ... can we find the linear operator with particular solution given http://en.wikipedia.org/wiki/Variation_of_parameters

OpenStudy (experimentx):

why do you have three particular solutions first of all? y=x , y=e^x+x , y=2e^x+x+1

OpenStudy (experimentx):

are these all particular solution of superposition of different inputs or .... two of them is part of solution of homogeneous equation?

OpenStudy (anonymous):

We haven't talked about variations of parameters though...

OpenStudy (anonymous):

Doesn't say anything about that. Those yp's were just given.

OpenStudy (anonymous):

Three particular solutions of a certain second order linear equation Ly=f are y=x, y=e^x+x, y=2e^x+1+x. What is the general solution?

OpenStudy (anonymous):

That's the complete problem.

OpenStudy (experimentx):

and f is also not given!! ... probably your answer is \[ y_h + y_1 + y_2 + y_3\] what level do you study man??

OpenStudy (anonymous):

College? I am currently looking into using wronskians with three functions. Maybe that would help me.

OpenStudy (anonymous):

The book we're using is from 1960's!!

OpenStudy (anonymous):

It doesn't give a lot of examples to work with before giving exercises.

OpenStudy (experimentx):

yeah i know you don't encounter these until you second year in university ... extracting info out of solution ... i've not tried yet.

OpenStudy (experimentx):

I don't know much about wronskian but i know it is used in method variation of parameters to find particular solution. also to check linear dependence between solutions.

OpenStudy (anonymous):

Well, I really appreciate your help man. At least I know I just need to add up all those yps.

OpenStudy (anonymous):

http://www.math.ucdavis.edu/~dragon/22B07/wronskian.pdf <= Looks like this would lead me to the right track

OpenStudy (experimentx):

well ... as far as i know ...and thought, |dw:1349634268921:dw| I thought in this way.

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