Logs - Simplifying with arbitrary constant I've tried simplifying my answer, which has a constant in it. I would like to know if I am on the right track: ln(y) = -(x^2)/(2y^2) + C C can be considered as an Arbitrary Constant. From this we can say that: ln(k)=C

ritualizi6zk

ritualizi6zk

Answered question

2022-11-03

Logs - Simplifying with arbitrary constant
I've tried simplifying my answer, which has a constant in it. I would like to know if I am on the right track:
ln ( y ) = x 2 2 y 2 + C
C can be considered as an Arbitrary Constant. From this we can say that:
ln ( k ) = C
We can further simplify the above to the following:
y = k . e x 2 2 y 2
Would this be correct?

Answer & Explanation

jennasyliang4tr

jennasyliang4tr

Beginner2022-11-04Added 15 answers

Yes, this is correct. To me, it is simpler to argue that
y = exp ( x 2 2 y 2 + C ) = e C exp ( x 2 / ( 2 y 2 ) ) = K e x 2 / ( 2 y 2 ) ,
where K is some constant since C is a constant.
Aryanna Fisher

Aryanna Fisher

Beginner2022-11-05Added 6 answers

We have,
ln ( y ) = x 2 2 y 2 + C
if ln ( k ) = C then
ln ( y ) = x 2 2 y 2 + ln ( k )
x 2 2 y 2 = ln ( y ) + ln ( k )
x 2 2 y 2 = ln ( k y )
k y = e x 2 2 y 2
y k = e x 2 2 y 2
y = k e x 2 2 y 2
Your answer is correct.

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