anonymous
  • anonymous
What is the integral of y/x dx?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

amistre64
  • amistre64
we can change this a bit: dy = (y/x) dx or dy/y = dx/x and integrate (S) both sides. (S) (1/y) dy = (S) (1/x) dx ln(y) = ln(x) + C e^(ln(y)) = e^(ln(x) + C) y=e^(ln(x)) * e^C y=Cx This is my best guess :)
Gina
  • Gina
yes you right
amistre64
  • amistre64
yay!! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Well…really, \[\int\limits(y/x)dx\]means to integrate with respect to x, treating everything else as a constant. Thus, \[y \int\limits(1/x)=y \ln \left| x \right|+C.\]Just sayin'.
amistre64
  • amistre64
I would love to agree with you carly; but, ... If we undo that we get: d(f(x))/dx = d(y ln(x))/dx f'(x) = y*(1/x)*(dx/dx) + (dydx)*(ln(x)) f'(x) = y/x + y' ln(x). ; which is not the same as (y/x) Right?
anonymous
  • anonymous
What is the derivative of ln x? Now what's that times a constant y, hmm?
amistre64
  • amistre64
But the "y" is not a constant; it is an implicit function of "x". If "y" is an independant variable of f(x,y) then yes, we would hold "y" as a constant to find the rate of change of f(x,y) with respect to "x".

Looking for something else?

Not the answer you are looking for? Search for more explanations.