I've tried simplifying my answer, which has a constant in it. I would like to know if I am on the right track: $$ \ln(y) = -{x^2\over 2y^2} + C $$ 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: $$ y = k.e^\frac{-x^2}{2y^2} $$
Would this be correct?
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We have, $$ \ln(y) = -{x^2\over 2y^2} + C $$ if $\ln(k) = C$ then $$ \ln(y) = -{x^2\over 2y^2} + \ln (k) $$ $$\implies \frac{x^2}{2y^2}=-\ln (y)+\ln(k)$$ $$\implies \frac{x^2}{2y^2}=\ln\left(\frac{k}{y}\right)$$ $$\implies \frac{k}{y}=e^{\frac{x^2}{2y^2}}$$ $$\implies \frac{y}{k}=e^{-\frac{x^2}{2y^2}}$$ $$\implies y=k\cdot e^{-\frac{x^2}{2y^2}}$$ Your answer is correct.
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Given that $C$ is a constant (meaning it does not change with any variable change), we can indeed substitute $C=\ln(k)$ for some other constant $k$. Though this constant is related to $C$, it is still a constant - because $C$ is in no way related to a variable, $C$ will not change, and neither will $k$. Your simplified form is thus correct.
Yes, this is correct. To me, it is simpler to argue that $$ y = \exp \left( -\frac{x^2}{2y^2}+C \right) = e^C \exp \left( -x^2/(2y^2) \right) = K e^{-x^2/(2y^2)}, $$ where $K$ is some constant since $C$ is a constant.