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? )

Question

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? )

liizzyliizz · Answer

@amistre64  seem ok? or no?

amistre64 · Answer

$$ (x^2-3) dy/dx = xy$$
$$ (x^2-3) dy = xy\ dx$$
$$ y^{-1} dy = x (x^2-3)^{-1}\ dx$$
$$ ln(y) = \frac12ln(x^2-3)+C$$

amistre64 · Answer

now we get to play with an e :)

$$\Large e^{ln(y) = \frac12ln(x^2-3)+C}$$
$$y = \large e^{ln(x^2-3)^{1/2}+C}$$
$$y = \large (x^2-3)^{1/2}*e^C$$
$$y = C(x^2-3)^{1/2}$$

amistre64 · Answer

i think yours got off in the e-ing the sides

liizzyliizz · Answer

I messed up the e^c at the end.. I was heading in the right direction though lol

amistre64 · Answer

yeah, you were :)  practice makes almost perfect tho lol

liizzyliizz · Answer

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 · Answer

$$n\ ln(a)=ln(a^n)$$
$$e^{ln(a^n)}=a^n$$

amistre64 · Answer

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 · Answer

if that makes sense lol

liizzyliizz · Answer

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 · Answer

no, since you e-ed out the ln but retained the 1/2 yours is a different beast

liizzyliizz · Answer

Oh ok  I see