liizzyliizz
  • liizzyliizz
ok so for the general solution of (x^2-3) dy/dx = xy I got y= (|x^2-3|)/2 +C (is this right or wrong? )
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
liizzyliizz
  • liizzyliizz
@amistre64 seem ok? or no?
amistre64
  • amistre64
\[ (x^2-3) dy/dx = xy\] \[ (x^2-3) dy = xy\ dx\] \[ y^{-1} dy = x (x^2-3)^{-1}\ dx\] \[ ln(y) = \frac12ln(x^2-3)+C\]
amistre64
  • amistre64
now we get to play with an e :) \[\Large e^{ln(y) = \frac12ln(x^2-3)+C}\] \[y = \large e^{ln(x^2-3)^{1/2}+C}\] \[y = \large (x^2-3)^{1/2}*e^C\] \[y = C(x^2-3)^{1/2}\]

Looking for something else?

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

More answers

amistre64
  • amistre64
i think yours got off in the e-ing the sides
liizzyliizz
  • liizzyliizz
I messed up the e^c at the end.. I was heading in the right direction though lol
amistre64
  • amistre64
yeah, you were :) practice makes almost perfect tho lol
liizzyliizz
  • liizzyliizz
the only thing I do not have that you do is the ^1/2 at the end, which has me a little thrown off. o_O
amistre64
  • amistre64
\[n\ ln(a)=ln(a^n)\] \[e^{ln(a^n)}=a^n\]
amistre64
  • amistre64
you have to wrap up that 1/2 that pops into the inegration, into the LN to get rid of the LN with an E
amistre64
  • amistre64
if that makes sense lol
liizzyliizz
  • liizzyliizz
It somewhat makes sense to me. lol I see what you're doing but i wonder if what I have would be equivalent to your answer lol
amistre64
  • amistre64
no, since you e-ed out the ln but retained the 1/2 yours is a different beast
liizzyliizz
  • liizzyliizz
Oh ok I see

Looking for something else?

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