Form of General Solution Regarding x

[Jason76]

In this example, x is on the top, but could it also be on the bottom?

\(\displaystyle \dfrac{x}{y} = -3\ln|x| + C_{1}\)

Now attempting to isolate y on the left side. So we do a reciprocal on both sides:

\(\displaystyle \dfrac{y}{x} = \dfrac{1}{-3\ln|x| + C_{1}}\)

\(\displaystyle y = \dfrac{x}{-3\ln|x| + C_{1}}\)

 

[HallsofIvy]

What are you trying to do? If you want to solve for y, to be able to write y= f(x), then, yes, to go from "y/x" to "y" you have to multiply both sides by x which gives x in the numerator. I'm not sure what you mean by "could it also be on the bottom?" If you mean simply "can we have x in the denominator as well as in the numerator", yes, of course. You can have very complicated functions.

 

[Jason76]

What are you trying to do? If you want to solve for y, to be able to write y= f(x), then, yes, to go from "y/x" to "y" you have to multiply both sides by x which gives x in the numerator. I'm not sure what you mean by "could it also be on the bottom?" If you mean simply "can we have x in the denominator as well as in the numerator", yes, of course. You can have very complicated functions.

My mistake. If you eliminate x without doing the reciprocal, then you have 1 over y. Later, you have to do a reciprocal anyways (to get it into proper form), Therefore, you end up with x on the top again.

\(\displaystyle \dfrac{x}{y} = -3\ln|x| + c_{1}\)

\(\displaystyle \dfrac{1}{y} = \dfrac{-3\ln|x| + c_{1}}{x}\)

So reciprocal is still required:

\(\displaystyle y = \dfrac{x}{-3\ln|x| + c_{1}}\)