# Normal distribution differential equation

The differential equation having a normal distribution as its solution is

 (60)

This dynamic will be dictated by the distance from mean, the slope  will be positive before mean and negative after mean and y will asymptotically reach 0.

since

 (61)
 (62)
 (63)

It can be shown that the optimums of change in Y happen at sigma far from the mean. In other words the maximum fall rate for the probability happens at x=+ – sigma

${y}'=y(\mu&space;-x)/\sigma&space;^{2}$

${y}''={y}'(\frac{\mu&space;-x)}{\sigma&space;^{2}}+y({\frac{\mu&space;-x}{\sigma&space;^{2}}})'=0$

$\frac{y(\mu&space;-x)}{\sigma&space;^{2}}(\frac{(\mu&space;-x)}{\sigma&space;^{2}})-\frac{y}{\sigma&space;^{2}}=0$

$y(\frac{(\mu&space;-x)^{2}-\sigma&space;^{2}}{\sigma^{4}&space;})=0$

$(\mu&space;-x)^{2}-\sigma&space;^{2}}=0$

$(\mu&space;-x)=\sigma$

This equation has been generalized to yield more complicated distributions which are named using the so-called Pearson system.

The normal distribution is also a special case of the chi-squared distribution, since making the substitution

 (64)

gives

 (65) (66)

Now, the real line is mapped onto the half-infinite interval by this transformation, so an extra factor of 2 must be added to , transforming into

 (67)