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solve the diffrential equation:
xy' = x(e^(-y/x)) + y
answer:
y = xln|lncx|
problem:
i don't get it
can anyone explain
xy' = x(e^(-y/x)) + y
answer:
y = xln|lncx|
problem:
i don't get it
can anyone explain
That would indeed be my next stepyeah i get that... but then i don't know what to do... do we substitute y=vx???
It is indeed. In general ln(a)+ln(b)=ln(ab).lnC + lnx is lnCx yeah?
What are the last few steps leading up to your answer? Without seeing your work, it is difficult to say where your error is. Though I am inclined to say that you are probably not taking the natural log of the entire right hand side. That is usually where my error is.ok
no matter what i do i'm not getting that answer! and its not supposed to a hard question!!
i get this:
y = x lnCx
No problemyes yes yes! i see
cheers!